Introductory Statistics is designed for the one-semester, introduction to statistics course and is geared toward students majoring in fields other than math or engineering. This text assumes students have been exposed to intermediate algebra, and it focuses on the applications of statistical knowledge rather than the theory behind it.
The foundation of this textbook is Collaborative Statistics, by Barbara Illowsky and Susan Dean. Additional topics, examples, and ample opportunities for practice have been added to each chapter. The development choices for this textbook were made with the guidance of many faculty members who are deeply involved in teaching this course. These choices led to innovations in art, terminology, and practical applications, all with a goal of increasing relevance and accessibility for students. We strove to make the discipline meaningful, so that students can draw from it a working knowledge that will enrich their future studies and help them make sense of the world around them.
Chapter 1 Sampling and Data
Introductory Statistics was conceived and written to fit a particular topical sequence, but it can be used flexibly to accommodate other course structures. One such potential structure, which will fit reasonably well with the textbook content, is provided. Please consider, however, that the chapters were not written to be completely independent, and that the proposed alternate sequence should be carefully considered for student preparation and textual consistency.
Chapter 1 Sampling and Data
These innovative activities were developed by Barbara Illowsky and Susan Dean in order to offer students the experience of designing, implementing, and interpreting statistical analyses. They are drawn from actual experiments and data-gathering processes, and offer a unique hands-on and collaborative experience. The labs provide a foundation for further learning and classroom interaction that will produce a meaningful application of statistics.
Statistics Labs appear at the end of each chapter, and begin with student learning outcomes, general estimates for time on task, and any global implementation notes. Students are then provided step-by-step guidance, including sample data tables and calculation prompts. The detailed assistance will help the students successfully apply the concepts in the text and lay the groundwork for future collaborative or individual work.
| Barbara Illowsky | De Anza College |
| Susan Dean | De Anza College |
| Abdulhamid Sukar | Cameron University |
| Abraham Biggs | Broward Community College |
| Adam Pennell | Greensboro College |
| Alexander Kolovos | |
| Andrew Wiesner | Pennsylvania State University |
| Ann Flanigan | Kapiolani Community College |
| Benjamin Ngwudike | Jackson State University |
| Birgit Aquilonius | West Valley College |
| Bryan Blount | Kentucky Wesleyan College |
| Carol Olmstead | De Anza College |
| Carol Weideman | St. Petersburg College |
| Charles Ashbacher | Upper Iowa University, Cedar Rapids |
| Charles Klein | De Anza College |
| Cheryl Wartman | University of Prince Edward Island |
| Cindy Moss | Skyline College |
| Daniel Birmajer | Nazareth College |
| David Bosworth | Hutchinson Community College |
| David French | Tidewater Community College |
| Dennis Walsh | Middle Tennessee State University |
| Diane Mathios | De Anza College |
| Ernest Bonat | Portland Community College |
| Frank Snow | De Anza College |
| George Bratton | University of Central Arkansas |
| Inna Grushko | De Anza College |
| Janice Hector | De Anza College |
| Javier Rueda | De Anza College |
| Jeffery Taub | Maine Maritime Academy |
| Jim Helmreich | Marist College |
| Jim Lucas | De Anza College |
| Jing Chang | College of Saint Mary |
| John Thomas | College of Lake County |
| Jonathan Oaks | Macomb Community College |
| Kathy Plum | De Anza College |
| Larry Green | Lake Tahoe Community College |
| Laurel Chiappetta | University of Pittsburgh |
| Lenore Desilets | De Anza College |
| Lisa Markus | De Anza College |
| Lisa Rosenberg | Elon University |
| Lynette Kenyon | Collin County Community College |
| Mark Mills | Central College |
| Mary Jo Kane | De Anza College |
| Mary Teegarden | San Diego Mesa College |
| Matthew Einsohn | Prescott College |
| Mel Jacobsen | Snow College |
| Michael Greenwich | College of Southern Nevada |
| Miriam Masullo | SUNY Purchase |
| Mo Geraghty | De Anza College |
| Nydia Nelson | St. Petersburg College |
| Philip J. Verrecchia | York College of Pennsylvania |
| Robert Henderson | Stephen F. Austin State University |
| Robert McDevitt | Germanna Community College |
| Roberta Bloom | De Anza College |
| Rupinder Sekhon | De Anza College |
| Sara Lenhart | Christopher Newport University |
| Sarah Boslaugh | Kennesaw State University |
| Sheldon Lee | Viterbo University |
| Sheri Boyd | Rollins College |
| Sudipta Roy | Kankakee Community College |
| Travis Short | St. Petersburg College |
| Valier Hauber | De Anza College |
| Vladimir Logvenenko | De Anza College |
| Wendy Lightheart | Lane Community College |
| Yvonne Sandoval | Pima Community College |
By the end of this chapter, the student should be able to:
You are probably asking yourself the question, "When and where will I use statistics?" If you read any newspaper, watch television, or use the Internet, you will see statistical information. There are statistics about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or watch a television news program, you are given sample information. With this information, you may make a decision about the correctness of a statement, claim, or "fact." Statistical methods can help you make the "best educated guess."
Since you will undoubtedly be given statistical information at some point in your life, you need to know some techniques for analyzing the information thoughtfully. Think about buying a house or managing a budget. Think about your chosen profession. The fields of economics, business, psychology, education, biology, law, computer science, police science, and early childhood development require at least one course in statistics.
Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what "good" data can be distinguished from "bad."
The science of statistics deals with the collection, analysis, interpretation, and presentation of data. We see and use data in our everyday lives.
In your classroom, try this exercise. Have class members write down the average time (in hours, to the nearest half-hour) they sleep per night. Your instructor will record the data. Then create a simple graph (called a dot plot) of the data. A dot plot consists of a number line and dots (or points) positioned above the number line. For example, consider the following data:
5 5.5 6 6 6 6.5 6.5 6.5 6.5 7 7 8 8 9
The dot plot for this data would be as follows:
Does your dot plot look the same as or different from the example? Why? If you did the same example in an English class with the same number of students, do you think the results would be the same? Why or why not?
Where do your data appear to cluster? How might you interpret the clustering?
The questions above ask you to analyze and interpret your data. With this example, you have begun your study of statistics.
In this course, you will learn how to organize and summarize data. Organizing and summarizing data is called descriptive statistics. Two ways to summarize data are by graphing and by using numbers (for example, finding an average). After you have studied probability and probability distributions, you will use formal methods for drawing conclusions from "good" data. The formal methods are called inferential statistics. Statistical inference uses probability to determine how confident we can be that our conclusions are correct.
Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful examination of the data. You will encounter what will seem to be too many mathematical formulas for interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to gain an understanding of your data. The calculations can be done using a calculator or a computer. The understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more confident in the decisions you make in life.
Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of an event occurring. For example, if you toss a fair coin four times, the outcomes may not be two heads and two tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails. The expected theoretical probability of heads in any one toss is or 0.5. Even though the outcomes of a few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012 heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction is equal to 0.498 which is very close to 0.5, the expected probability.
The theory of probability began with the study of games of chance such as poker. Predictions take the form of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of return on a client's investments. You might use probability to decide to buy a lottery ticket or not. In your study of statistics, you will use the power of mathematics through probability calculations to analyze and interpret your data.
In statistics, we generally want to study a population. You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample. The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population.
Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16 ounce can contains 16 ounces of carbonated drink.
From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter. A parameter is a number that is a property of the population. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter.
One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample. We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter.
A variable, notated by capital letters such as X and Y, is a characteristic of interest for each person or thing in a population. Variables may be numerical or categorical. Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category. If we let X equal the number of points earned by one math student at the end of a term, then X is a numerical variable. If we let Y be a person's party affiliation, then some examples of Y include Republican, Democrat, and Independent. Y is a categorical variable. We could do some math with values of X (calculate the average number of points earned, for example), but it makes no sense to do math with values of Y (calculating an average party affiliation makes no sense).
Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value.
Two words that come up often in statistics are mean and proportion. If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is and the proportion of women students is . Mean and proportion are discussed in more detail in later chapters.
The words "mean" and "average" are often used interchangeably. The substitution of one word for the other is common practice. The technical term is "arithmetic mean," and "average" is technically a center location. However, in practice among non-statisticians, "average" is commonly accepted for "arithmetic mean."
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly survey 100 first year students at the college. Three of those students spent $150, $200, and $225, respectively.
The population is all first year students attending ABC College this term.
The sample could be all students enrolled in one section of a beginning statistics course at ABC College (although this sample may not represent the entire population).
The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term.
The statistic is the average (mean) amount of money spent (excluding books) by first year college students in the sample.
The variable could be the amount of money spent (excluding books) by one first year student. Let X = the amount of money spent (excluding books) by one first year student attending ABC College.
The data are the dollar amounts spent by the first year students. Examples of the data are $150, $200, and $225.
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65, $75, and $95, respectively.
The population is all families with children attending Knoll Academy.
The sample is a random selection of 100 families with children attending Knoll Academy.
The parameter is the average (mean) amount of money spent on school uniforms by families with children at Knoll Academy.
The statistic is the average (mean) amount of money spent on school uniforms by families in the sample.
The variable is the amount of money spent by one family. Let X = the amount of money spent on school uniforms by one family with children attending Knoll Academy.
The data are the dollar amounts spent by the families. Examples of the data are $65, $75, and $95.
Determine what the key terms refer to in the following study.
A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below.
1._____ Population 2._____ Statistic 3._____ Parameter 4._____ Sample 5._____ Variable 6._____ Data
1. f 2. g 3. e 4. d 5. b 6. c
Determine what the key terms refer to in the following study.
As part of a study designed to test the safety of automobiles, the National Transportation Safety Board collected and reviewed data about the effects of an automobile crash on test dummies. Here is the criterion they used:
| Speed at which Cars Crashed | Location of “drive” (i.e. dummies) |
| 35 miles/hour | Front Seat |
Cars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars.
The population is all cars containing dummies in the front seat.
The sample is the 75 cars, selected by a simple random sample.
The parameter is the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population.
The statistic is proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample.
The variable X = the number of driver dummies (if they had been real people) who would have suffered head injuries.
The data are either: yes, had head injury, or no, did not.
Determine what the key terms refer to in the following study.
An insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit.
The population is all medical doctors listed in the professional directory.
The parameter is the proportion of medical doctors who have been involved in one or more malpractice suits in the population.
The sample is the 500 doctors selected at random from the professional directory.
The statistic is the proportion of medical doctors who have been involved in one or more malpractice suits in the sample.
The variable X = the number of medical doctors who have been involved in one or more malpractice suits.
The data are either: yes, was involved in one or more malpractice lawsuits, or no, was not.
Do the following exercise collaboratively with up to four people per group. Find a population, a sample, the parameter, the statistic, a variable, and data for the following study: You want to determine the average (mean) number of glasses of milk college students drink per day. Suppose yesterday, in your English class, you asked five students how many glasses of milk they drank the day before. The answers were 1, 0, 1, 3, and 4 glasses of milk.
Data may come from a population or from a sample. Small letters like or generally are used to represent data values. Most data can be put into the following categories:
Qualitative data are the result of categorizing or describing attributes of a population. Hair color, blood type, ethnic group, the car a person drives, and the street a person lives on are examples of qualitative data. Qualitative data are generally described by words or letters. For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Researchers often prefer to use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or blood type.
Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes of a population. Amount of money, pulse rate, weight, number of people living in your town, and number of students who take statistics are examples of quantitative data. Quantitative data may be either discrete or continuous.
All data that are the result of counting are called quantitative discrete data. These data take on only certain numerical values. If you count the number of phone calls you receive for each day of the week, you might get values such as zero, one, two, or three.
All data that are the result of measuring are quantitative continuous data assuming that we can measure accurately. Measuring angles in radians might result in such numbers as , , , , , and so on. If you and your friends carry backpacks with books in them to school, the numbers of books in the backpacks are discrete data and the weights of the backpacks are continuous data.
The data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data.
The data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this?
quantitative discrete data
The data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data because weights are measured.
The data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this?
quantitative continuous data
You go to the supermarket and purchase three cans of soup (19 ounces) tomato bisque, 14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts and peanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), and two desserts (16 ounces Cherry Garcia ice cream and two pounds (32 ounces chocolate chip cookies).
Name data sets that are quantitative discrete, quantitative continuous, and qualitative.
One Possible Solution:
Try to identify additional data sets in this example.
The data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative data.
The data are the colors of houses. You sample five houses. The colors of the houses are white, yellow, white, red, and white. What type of data is this?
qualitative data
You may collect data as numbers and report it categorically. For example, the quiz scores for each student are recorded throughout the term. At the end of the term, the quiz scores are reported as A, B, C, D, or F.
Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words "the number of."
Items a, e, f, k, and l are quantitative discrete; items d, j, and n are quantitative continuous; items b, c, g, h, i, and m are qualitative.
Determine the correct data type (quantitative or qualitative) for the number of cars in a parking lot. Indicate whether quantitative data are continuous or discrete.
quantitative discrete
A statistics professor collects information about the classification of her students as freshmen, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart Figure 1.3. What type of data does this graph show?
This pie chart shows the students in each year, which is qualitative data.
The registrar at State University keeps records of the number of credit hours students complete each semester. The data he collects are summarized in the histogram. The class boundaries are 10 to less than 13, 13 to less than 16, 16 to less than 19, 19 to less than 22, and 22 to less than 25.
A histogram is used to display quantitative data: the numbers of credit hours completed. Because students can complete only a whole number of hours (no fractions of hours allowed), this data is quantitative discrete.
Below are tables comparing the number of part-time and full-time students at De Anza College and Foothill College enrolled for the spring 2010 quarter. The tables display counts (frequencies) and percentages or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is compared to De Anza College.
| De Anza College | Foothill College | |||||
|---|---|---|---|---|---|---|
| Number | Percent | Number | Percent | |||
| Full-time | 9,200 | 40.9% | Full-time | 4,059 | 28.6% | |
| Part-time | 13,296 | 59.1% | Part-time | 10,124 | 71.4% | |
| Total | 22,496 | 100% | Total | 14,183 | 100% | |
Tables are a good way of organizing and displaying data. But graphs can be even more helpful in understanding the data. There are no strict rules concerning which graphs to use. Two graphs that are used to display qualitative data are pie charts and bar graphs.
In a pie chart, categories of data are represented by wedges in a circle and are proportional in size to the percent of individuals in each category.
In a bar graph, the length of the bar for each category is proportional to the number or percent of individuals in each category. Bars may be vertical or horizontal.
A Pareto chart consists of bars that are sorted into order by category size (largest to smallest).
Look at Figure 1.5 and Figure 1.6 and determine which graph (pie or bar) you think displays the comparisons better.
It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the data. We might make different choices of what we think is the “best” graph depending on the data and the context. Our choice also depends on what we are using the data for.
Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages add to more than 100% because students can be in more than one category. A bar graph is appropriate to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the percentages added to less than 100%.
| Characteristic/Category | Percent |
|---|---|
| Full-Time Students | 40.9% |
| Students who intend to transfer to a 4-year educational institution | 48.6% |
| Students under age 25 | 61.0% |
| TOTAL | 150.5% |
The table displays Ethnicity of Students but is missing the "Other/Unknown" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart.
| Frequency | Percent | |
|---|---|---|
| Asian | 8,794 | 36.1% |
| Black | 1,412 | 5.8% |
| Filipino | 1,298 | 5.3% |
| Hispanic | 4,180 | 17.1% |
| Native American | 146 | 0.6% |
| Pacific Islander | 236 | 1.0% |
| White | 5,978 | 24.5% |
| TOTAL | 22,044 out of 24,382 | 90.4% out of 100% |
The following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us.
This particular bar graph in Figure 1.9 can be difficult to understand visually. The graph in Figure 1.10 is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret.
The following pie charts have the “Other/Unknown” category included (since the percentages must add to 100%). The chart in Figure 1.11 is organized by the size of each wedge, which makes it a more visually informative graph than the unsorted, alphabetical graph in Figure 1.11.
Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling. In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons. The easiest method to describe is called a simple random sample. Any group of n individuals is equally likely to be chosen by any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus class, which has 31 members not including Lisa. To choose a simple random sample of size three from the other members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out three names. A more technological way is for Lisa to first list the last names of the members of her class together with a two-digit number, as in Table 1.5:
| ID | Name | ID | Name | ID | Name |
|---|---|---|---|---|---|
| 00 | Anselmo | 11 | King | 21 | Roquero |
| 01 | Bautista | 12 | Legeny | 22 | Roth |
| 02 | Bayani | 13 | Lundquist | 23 | Rowell |
| 03 | Cheng | 14 | Macierz | 24 | Salangsang |
| 04 | Cuarismo | 15 | Motogawa | 25 | Slade |
| 05 | Cuningham | 16 | Okimoto | 26 | Stratcher |
| 06 | Fontecha | 17 | Patel | 27 | Tallai |
| 07 | Hong | 18 | Price | 28 | Tran |
| 08 | Hoobler | 19 | Quizon | 29 | Wai |
| 09 | Jiao | 20 | Reyes | 30 | Wood |
| 10 | Khan |
Lisa can use a table of random numbers (found in many statistics books and mathematical handbooks), a calculator, or a computer to generate random numbers. For this example, suppose Lisa chooses to generate random numbers from a calculator. The numbers generated are as follows:
0.94360 0.99832 0.14669 0.51470 0.40581 0.73381 0.04399
Lisa reads two-digit groups until she has chosen three class members (that is, she reads 0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could have generated more random numbers.
The random numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However the third random number, 0.14669, contains 14 (the fourth random number also contains 14), the fifth random number contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist of Marcierz, Cuningham, and Cuarismo.
To generate random numbers:
Note: randInt(0, 30, 3) will generate 3 random numbers.
Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster sample, and the systematic sample.
To choose a stratified sample, divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample.
To choose a cluster sample, divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample.
To choose a systematic sample, randomly select a starting point and take every nth piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method.
A type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others.
Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents.
True random sampling is done with replacement. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low.
In a college population of 10,000 people, suppose you want to pick a sample of 1,000 randomly for a survey. For any particular sample of 1,000, if you are sampling with replacement,
If you are sampling without replacement,
Compare the fractions 999/10,000 and 999/9,999. For accuracy, carry the decimal answers to four decimal places. To four decimal places, these numbers are equivalent (0.0999).
Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. For example, if the population is 25 people, the sample is ten, and you are sampling with replacement for any particular sample, then the chance of picking the first person is ten out of 25, and the chance of picking a different second person is nine out of 25 (you replace the first person).
If you sample without replacement, then the chance of picking the first person is ten out of 25, and then the chance of picking the second person (who is different) is nine out of 24 (you do not replace the first person).
Compare the fractions 9/25 and 9/24. To four decimal places, 9/25 = 0.3600 and 9/24 = 0.3750. To four decimal places, these numbers are not equivalent.
When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors. A defective counting device can cause a nonsampling error.
In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error.
In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied.
A study is done to determine the average tuition that San Jose State undergraduate students pay per semester. Each student in the following samples is asked how much tuition he or she paid for the Fall semester. What is the type of sampling in each case?
a. stratified; b. systematic; c. simple random; d. cluster; e. convenience
You are going to use the random number generator to generate different types of samples from the data.
This table displays six sets of quiz scores (each quiz counts 10 points) for an elementary statistics class.
| #1 | #2 | #3 | #4 | #5 | #6 |
|---|---|---|---|---|---|
| 5 | 7 | 10 | 9 | 8 | 3 |
| 10 | 5 | 9 | 8 | 7 | 6 |
| 9 | 10 | 8 | 6 | 7 | 9 |
| 9 | 10 | 10 | 9 | 8 | 9 |
| 7 | 8 | 9 | 5 | 7 | 4 |
| 9 | 9 | 9 | 10 | 8 | 7 |
| 7 | 7 | 10 | 9 | 8 | 8 |
| 8 | 8 | 9 | 10 | 8 | 8 |
| 9 | 7 | 8 | 7 | 7 | 8 |
| 8 | 8 | 10 | 9 | 8 | 7 |
Instructions: Use the Random Number Generator to pick samples.
Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f.convenience
Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
A high school principal polls 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors regarding policy changes for after school activities.
stratified
If we were to examine two samples representing the same population, even if we used random sampling methods for the samples, they would not be exactly the same. Just as there is variation in data, there is variation in samples. As you become accustomed to sampling, the variability will begin to seem natural.
Suppose ABC College has 10,000 part-time students (the population). We are interested in the average amount of money a part-time student spends on books in the fall term. Asking all 10,000 students is an almost impossible task.
Suppose we take two different samples.
First, we use convenience sampling and survey ten students from a first term organic chemistry class. Many of these students are taking first term calculus in addition to the organic chemistry class. The amount of money they spend on books is as follows:
$128 $87 $173 $116 $130 $204 $147 $189 $93 $153
The second sample is taken using a list of senior citizens who take P.E. classes and taking every fifth senior citizen on the list, for a total of ten senior citizens. They spend:
$50 $40 $36 $15 $50 $100 $40 $53 $22 $22
It is unlikely that any student is in both samples.
a. Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population?
a. No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are also taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average parttime student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample.
b. Since these samples are not representative of the entire population, is it wise to use the results to describe the entire population?
b. No. For these samples, each member of the population did not have an equally likely chance of being chosen.
Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education, art, and early childhood development. (We assume that these are the only disciplines in which part-time students at ABC College are enrolled and that an equal number of part-time students are enrolled in each of the disciplines.) Each student is chosen using simple random sampling. Using a calculator, random numbers are generated and a student from a particular discipline is selected if he or she has a corresponding number. The students spend the following amounts:
$180 $50 $150 $85 $260 $75 $180 $200 $200 $150
c. Is the sample biased?
c. The sample is unbiased, but a larger sample would be recommended to increase the likelihood that the sample will be close to representative of the population. However, for a biased sampling technique, even a large sample runs the risk of not being representative of the population.
Students often ask if it is "good enough" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes.
A local radio station has a fan base of 20,000 listeners. The station wants to know if its audience would prefer more music or more talk shows. Asking all 20,000 listeners is an almost impossible task.
The station uses convenience sampling and surveys the first 200 people they meet at one of the station’s music concert events. 24 people said they’d prefer more talk shows, and 176 people said they’d prefer more music.
Do you think that this sample is representative of (or is characteristic of) the entire 20,000 listener population?
The sample probably consists more of people who prefer music because it is a concert event. Also, the sample represents only those who showed up to the event earlier than the majority. The sample probably doesn’t represent the entire fan base and is probably biased towards people who would prefer music.
As a class, determine whether or not the following samples are representative. If they are not, discuss the reasons.
Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less than 16 ounces of liquid. In one study, eight 16 ounce cans were measured and produced the following amount (in ounces) of beverage:
15.8 16.1 15.2 14.8 15.8 15.9 16.0 15.5
Measurements of the amount of beverage in a 16-ounce can may vary because different people make the measurements or because the exact amount, 16 ounces of liquid, was not put into the cans. Manufacturers regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range.
Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the same purpose. This is completely natural. However, if two or more of you are taking the same data and get very different results, it is time for you and the others to reevaluate your data-taking methods and your accuracy.
It was mentioned previously that two or more samples from the same population, taken randomly, and having close to the same characteristics of the population will likely be different from each other. Suppose Doreen and Jung both decide to study the average amount of time students at their college sleep each night. Doreen and Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample will be different from Jung's sample. Even if Doreen and Jung used the same sampling method, in all likelihood their samples would be different. Neither would be wrong, however.
Think about what contributes to making Doreen’s and Jung’s samples different.
If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results (the average amount of time a student sleeps) might be closer to the actual population average. But still, their samples would be, in all likelihood, different from each other. This variability in samples cannot be stressed enough.
The size of a sample (often called the number of observations) is important. The examples you have seen in this book so far have been small. Samples of only a few hundred observations, or even smaller, are sufficient for many purposes. In polling, samples that are from 1,200 to 1,500 observations are considered large enough and good enough if the survey is random and is well done. You will learn why when you study confidence intervals.
Be aware that many large samples are biased. For example, call-in surveys are invariably biased, because people choose to respond or not.
Divide into groups of two, three, or four. Your instructor will give each group one six-sided die. Try this experiment twice. Roll one fair die (six-sided) 20 times. Record the number of ones, twos, threes, fours, fives, and sixes you get in Table 1.7 and Table 1.8 (“frequency” is the number of times a particular face of the die occurs):
| Face on Die | Frequency |
|---|---|
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 |
| Face on Die | Frequency |
|---|---|
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 |
Did the two experiments have the same results? Probably not. If you did the experiment a third time, do you expect the results to be identical to the first or second experiment? Why or why not?
Which experiment had the correct results? They both did. The job of the statistician is to see through the variability and draw appropriate conclusions.
We need to evaluate the statistical studies we read about critically and analyze them before accepting the results of the studies. Common problems to be aware of include
Once you have a set of data, you will need to organize it so that you can analyze how frequently each datum occurs in the set. However, when calculating the frequency, you may need to round your answers so that they are as precise as possible.
A simple way to round off answers is to carry your final answer one more decimal place than was present in the original data. Round off only the final answer. Do not round off any intermediate results, if possible. If it becomes necessary to round off intermediate results, carry them to at least twice as many decimal places as the final answer. For example, the average of the three quiz scores four, six, and nine is 6.3, rounded off to the nearest tenth, because the data are whole numbers. Most answers will be rounded off in this manner.
It is not necessary to reduce most fractions in this course. Especially in Probability Topics, the chapter on probability, it is more helpful to leave an answer as an unreduced fraction.
The way a set of data is measured is called its level of measurement. Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level):
Data that is measured using a nominal scale is qualitative. Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. For example, trying to classify people according to their favorite food does not make any sense. Putting pizza first and sushi second is not meaningful.
Smartphone companies are another example of nominal scale data. Some examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list and there is no agreed upon order. Some people may favor Apple but that is a matter of opinion. Nominal scale data cannot be used in calculations.
Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data.
Another example of using the ordinal scale is a cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations.
Data that is measured using the interval scale is similar to ordinal level data because it has a definite ordering but there is a difference between data. The differences between interval scale data can be measured though the data does not have a starting point.
Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not because, in both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0.
Interval level data can be used in calculations, but one type of comparison cannot be done. 80° C is not four times as hot as 20° C (nor is 80° F four times as hot as 20° F). There is no meaning to the ratio of 80 to 20 (or four to one).
Data that is measured using the ratio scale takes care of the ratio problem and gives you the most information. Ratio scale data is like interval scale data, but it has a 0 point and ratios can be calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points). The exams are machine-graded.
The data can be put in order from lowest to highest: 20, 68, 80, 92.
The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The smallest score is 0. So 80 is four times 20. The score of 80 is four times better than the score of 20.
Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows: 56332475235654435253.
Table 1.9 lists the different data values in ascending order and their frequencies.
| DATA VALUE | FREQUENCY |
|---|---|
| 2 | 3 |
| 3 | 5 |
| 4 | 3 |
| 5 | 6 |
| 6 | 2 |
| 7 | 1 |
A frequency is the number of times a value of the data occurs. According to Table 1.9, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.
A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.
| DATA VALUE | FREQUENCY | RELATIVE FREQUENCY |
|---|---|---|
| 2 | 3 | or 0.15 |
| 3 | 5 | or 0.25 |
| 4 | 3 | or 0.15 |
| 5 | 6 | or 0.30 |
| 6 | 2 | or 0.10 |
| 7 | 1 | or 0.05 |
The sum of the values in the relative frequency column of Table 1.10 is , or 1.
Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in Table 1.11.
| DATA VALUE | FREQUENCY | RELATIVE |
CUMULATIVE RELATIVE |
|---|---|---|---|
| 2 | 3 | or 0.15 | 0.15 |
| 3 | 5 | or 0.25 | 0.15 + 0.25 = 0.40 |
| 4 | 3 | or 0.15 | 0.40 + 0.15 = 0.55 |
| 5 | 6 | or 0.30 | 0.55 + 0.30 = 0.85 |
| 6 | 2 | or 0.10 | 0.85 + 0.10 = 0.95 |
| 7 | 1 | or 0.05 | 0.95 + 0.05 = 1.00 |
The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.
Because of rounding, the relative frequency column may not always sum to one, and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one.
Table 1.12 represents the heights, in inches, of a sample of 100 male semiprofessional soccer players.
| HEIGHTS |
FREQUENCY | RELATIVE |
CUMULATIVE |
|---|---|---|---|
| 59.95–61.95 | 5 | = 0.05 | 0.05 |
| 61.95–63.95 | 3 | = 0.03 | 0.05 + 0.03 = 0.08 |
| 63.95–65.95 | 15 | = 0.15 | 0.08 + 0.15 = 0.23 |
| 65.95–67.95 | 40 | = 0.40 | 0.23 + 0.40 = 0.63 |
| 67.95–69.95 | 17 | = 0.17 | 0.63 + 0.17 = 0.80 |
| 69.95–71.95 | 12 | = 0.12 | 0.80 + 0.12 = 0.92 |
| 71.95–73.95 | 7 | = 0.07 | 0.92 + 0.07 = 0.99 |
| 73.95–75.95 | 1 | = 0.01 | 0.99 + 0.01 = 1.00 |
| Total = 100 | Total = 1.00 |
The data in this table have been grouped into the following intervals:
This example is used again in Descriptive Statistics, where the method used to compute the intervals will be explained.
In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints.
From Table 1.12, find the percentage of heights that are less than 65.95 inches.
If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are 5 + 3 + 15 = 23 players whose heights are less than 65.95 inches. The percentage of heights less than 65.95 inches is then or 23%. This percentage is the cumulative relative frequency entry in the third row.
Table 1.13 shows the amount, in inches, of annual rainfall in a sample of towns.
| Rainfall (Inches) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 2.95–4.97 | 6 | = 0.12 | 0.12 |
| 4.97–6.99 | 7 | = 0.14 | 0.12 + 0.14 = 0.26 |
| 6.99–9.01 | 15 | = 0.30 | 0.26 + 0.30 = 0.56 |
| 9.01–11.03 | 8 | = 0.16 | 0.56 + 0.16 = 0.72 |
| 11.03–13.05 | 9 | = 0.18 | 0.72 + 0.18 = 0.90 |
| 13.05–15.07 | 5 | = 0.10 | 0.90 + 0.10 = 1.00 |
| Total = 50 | Total = 1.00 |
From Table 1.13, find the percentage of rainfall that is less than 9.01 inches.
0.56 or 56%
From Table 1.12, find the percentage of heights that fall between 61.95 and 65.95 inches.
Add the relative frequencies in the second and third rows: 0.03 + 0.15 = 0.18 or 18%.
From Table 1.13, find the percentage of rainfall that is between 6.99 and 13.05 inches.
0.30 + 0.16 + 0.18 = 0.64 or 64%
Use the heights of the 100 male semiprofessional soccer players in Table 1.12. Fill in the blanks and check your answers.
Remember, you count frequencies. To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row.
From Table 1.13, find the number of towns that have rainfall between 2.95 and 9.01 inches.
6 + 7 + 15 = 28 towns
In your class, have someone conduct a survey of the number of siblings (brothers and sisters) each student has. Create a frequency table. Add to it a relative frequency column and a cumulative relative frequency column. Answer the following questions:
Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are as follows: 2 5 7 3 2 10 18 15 20 7 10 18 5 12 13 12 4 5 10. Table 1.14 was produced:
| DATA | FREQUENCY | RELATIVE |
CUMULATIVE |
|---|---|---|---|
| 3 | 3 | 0.1579 | |
| 4 | 1 | 0.2105 | |
| 5 | 3 | 0.1579 | |
| 7 | 2 | 0.2632 | |
| 10 | 3 | 0.4737 | |
| 12 | 2 | 0.7895 | |
| 13 | 1 | 0.8421 | |
| 15 | 1 | 0.8948 | |
| 18 | 1 | 0.9474 | |
| 20 | 1 | 1.0000 |
Table 1.13 represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year?
Table 1.15 contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.
| Year | Total Number of Deaths |
|---|---|
| 2000 | 231 |
| 2001 | 21,357 |
| 2002 | 11,685 |
| 2003 | 33,819 |
| 2004 | 228,802 |
| 2005 | 88,003 |
| 2006 | 6,605 |
| 2007 | 712 |
| 2008 | 88,011 |
| 2009 | 1,790 |
| 2010 | 320,120 |
| 2011 | 21,953 |
| 2012 | 768 |
| Total | 823,356 |
Answer the following questions.
Table 1.16 contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.
| Year | Total Number of Crashes | Year | Total Number of Crashes |
|---|---|---|---|
| 1994 | 36,254 | 2004 | 38,444 |
| 1995 | 37,241 | 2005 | 39,252 |
| 1996 | 37,494 | 2006 | 38,648 |
| 1997 | 37,324 | 2007 | 37,435 |
| 1998 | 37,107 | 2008 | 34,172 |
| 1999 | 37,140 | 2009 | 30,862 |
| 2000 | 37,526 | 2010 | 30,296 |
| 2001 | 37,862 | 2011 | 29,757 |
| 2002 | 38,491 | Total | 653,782 |
| 2003 | 38,477 |
Answer the following questions.
Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data.
The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the explanatory variable. The affected variable is called the response variable. In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments. An experimental unit is a single object or individual to be measured.
You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention.
Additional variables that can cloud a study are called lurking variables. In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables.
The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted:
Results showed that believing one had taken the substance resulted in [performance] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment.1
When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group. This group is given a placebo treatment–a treatment that cannot influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, he does not know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded.
Researchers want to investigate whether taking aspirin regularly reduces the risk of heart attack. Four hundred men between the ages of 50 and 84 are recruited as participants. The men are divided randomly into two groups: one group will take aspirin, and the other group will take a placebo. Each man takes one pill each day for three years, but he does not know whether he is taking aspirin or the placebo. At the end of the study, researchers count the number of men in each group who have had heart attacks.
Identify the following values for this study: population, sample, experimental units, explanatory variable, response variable, treatments.
The population is men aged 50 to 84.
The Smell & Taste Treatment and Research Foundation conducted a study to investigate whether smell can affect learning. Subjects completed mazes multiple times while wearing masks. They completed the pencil and paper mazes three times wearing floral-scented masks, and three times with unscented masks. Participants were assigned at random to wear the floral mask during the first three trials or during the last three trials. For each trial, researchers recorded the time it took to complete the maze and the subject’s impression of the mask’s scent: positive, negative, or neutral.
A researcher wants to study the effects of birth order on personality. Explain why this study could not be conducted as a randomized experiment. What is the main problem in a study that cannot be designed as a randomized experiment?
The explanatory variable is birth order. You cannot randomly assign a person’s birth order. Random assignment eliminates the impact of lurking variables. When you cannot assign subjects to treatment groups at random, there will be differences between the groups other than the explanatory variable.
You are concerned about the effects of texting on driving performance. Design a study to test the response time of drivers while texting and while driving only. How many seconds does it take for a driver to respond when a leading car hits the brakes?
The widespread misuse and misrepresentation of statistical information often gives the field a bad name. Some say that “numbers don’t lie,” but the people who use numbers to support their claims often do.
A recent investigation of famous social psychologist, Diederik Stapel, has led to the retraction of his articles from some of the world’s top journals including Journal of Experimental Social Psychology, Social Psychology, Basic and Applied Social Psychology, British Journal of Social Psychology, and the magazine Science. Diederik Stapel is a former professor at Tilburg University in the Netherlands. Over the past two years, an extensive investigation involving three universities where Stapel has worked concluded that the psychologist is guilty of fraud on a colossal scale. Falsified data taints over 55 papers he authored and 10 Ph.D. dissertations that he supervised.
Stapel did not deny that his deceit was driven by ambition. But it was more complicated than that, he told me. He insisted that he loved social psychology but had been frustrated by the messiness of experimental data, which rarely led to clear conclusions. His lifelong obsession with elegance and order, he said, led him to concoct sexy results that journals found attractive. “It was a quest for aesthetics, for beauty—instead of the truth,” he said. He described his behavior as an addiction that drove him to carry out acts of increasingly daring fraud, like a junkie seeking a bigger and better high.2
The committee investigating Stapel concluded that he is guilty of several practices including:
Clearly, it is never acceptable to falsify data the way this researcher did. Sometimes, however, violations of ethics are not as easy to spot.
Researchers have a responsibility to verify that proper methods are being followed. The report describing the investigation of Stapel’s fraud states that, “statistical flaws frequently revealed a lack of familiarity with elementary statistics.”3 Many of Stapel’s co-authors should have spotted irregularities in his data. Unfortunately, they did not know very much about statistical analysis, and they simply trusted that he was collecting and reporting data properly.
Many types of statistical fraud are difficult to spot. Some researchers simply stop collecting data once they have just enough to prove what they had hoped to prove. They don’t want to take the chance that a more extensive study would complicate their lives by producing data contradicting their hypothesis.
Professional organizations, like the American Statistical Association, clearly define expectations for researchers. There are even laws in the federal code about the use of research data.
When a statistical study uses human participants, as in medical studies, both ethics and the law dictate that researchers should be mindful of the safety of their research subjects. The U.S. Department of Health and Human Services oversees federal regulations of research studies with the aim of protecting participants. When a university or other research institution engages in research, it must ensure the safety of all human subjects. For this reason, research institutions establish oversight committees known as Institutional Review Boards (IRB). All planned studies must be approved in advance by the IRB. Key protections that are mandated by law include the following:
These ideas may seem fundamental, but they can be very difficult to verify in practice. Is removing a participant’s name from the data record sufficient to protect privacy? Perhaps the person’s identity could be discovered from the data that remains. What happens if the study does not proceed as planned and risks arise that were not anticipated? When is informed consent really necessary? Suppose your doctor wants a blood sample to check your cholesterol level. Once the sample has been tested, you expect the lab to dispose of the remaining blood. At that point the blood becomes biological waste. Does a researcher have the right to take it for use in a study?
It is important that students of statistics take time to consider the ethical questions that arise in statistical studies. How prevalent is fraud in statistical studies? You might be surprised—and disappointed. There is a website (www.retractionwatch.com) dedicated to cataloging retractions of study articles that have been proven fraudulent. A quick glance will show that the misuse of statistics is a bigger problem than most people realize.
Vigilance against fraud requires knowledge. Learning the basic theory of statistics will empower you to analyze statistical studies critically.
Describe the unethical behavior in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected.
A researcher is collecting data in a community.
Describe the unethical behavior, if any, in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected.
A study is commissioned to determine the favorite brand of fruit juice among teens in California.
Class Time:
Names:
Movie SurveyAsk five classmates from a different class how many movies they saw at the theater last month. Do not include rented movies.
| ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ |
| ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ |
| ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ |
| ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ |
| ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___ |
Order the DataComplete the two relative frequency tables below using your class data.
| Number of Movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | |||
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 | |||
| 7+ |
| Number of Movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0–1 | |||
| 2–3 | |||
| 4–5 | |||
| 6–7+ |
Class Time:
Names:
In this lab, you will be asked to pick several random samples of restaurants. In each case, describe your procedure briefly, including how you might have used the random number generator, and then list the restaurants in the sample you obtained.
The following section contains restaurants stratified by city into columns and grouped horizontally by entree cost (clusters).
Restaurants Stratified by City and Entree Cost
| Entree Cost | Under $10 | $10 to under $15 | $15 to under $20 | Over $20 |
|---|---|---|---|---|
| San Jose | El Abuelo Taq, Pasta Mia, Emma’s Express, Bamboo Hut | Emperor’s Guard, Creekside Inn | Agenda, Gervais, Miro’s | Blake’s, Eulipia, Hayes Mansion, Germania |
| Palo Alto | Senor Taco, Olive Garden, Taxi’s | Ming’s, P.A. Joe’s, Stickney’s | Scott’s Seafood, Poolside Grill, Fish Market | Sundance Mine, Maddalena’s, Spago’s |
| Los Gatos | Mary’s Patio, Mount Everest, Sweet Pea’s, Andele Taqueria | Lindsey’s, Willow Street | Toll House | Charter House, La Maison Du Cafe |
| Mountain View | Maharaja, New Ma’s, Thai-Rific, Garden Fresh | Amber Indian, La Fiesta, Fiesta del Mar, Dawit | Austin’s, Shiva’s, Mazeh | Le Petit Bistro |
| Cupertino | Hobees, Hung Fu, Samrat, Panda Express | Santa Barb. Grill, Mand. Gourmet, Bombay Oven, Kathmandu West | Fontana’s, Blue Pheasant | Hamasushi, Helios |
| Sunnyvale | Chekijababi, Taj India, Full Throttle, Tia Juana, Lemon Grass | Pacific Fresh, Charley Brown’s, Cafe Cameroon, Faz, Aruba’s | Lion & Compass, The Palace, Beau Sejour | |
| Santa Clara | Rangoli, Armadillo Willy’s, Thai Pepper, Pasand | Arthur’s, Katie’s Cafe, Pedro’s, La Galleria | Birk’s, Truya Sushi, Valley Plaza | Lakeside, Mariani’s |
A Simple Random SamplePick a simple random sample of 15 restaurants.
| 1. __________ | 6. __________ | 11. __________ |
| 2. __________ | 7. __________ | 12. __________ |
| 3. __________ | 8. __________ | 13. __________ |
| 4. __________ | 9. __________ | 14. __________ |
| 5. __________ | 10. __________ | 15. __________ |
A Systematic SamplePick a systematic sample of 15 restaurants.
| 1. __________ | 6. __________ | 11. __________ |
| 2. __________ | 7. __________ | 12. __________ |
| 3. __________ | 8. __________ | 13. __________ |
| 4. __________ | 9. __________ | 14. __________ |
| 5. __________ | 10. __________ | 15. __________ |
A Stratified SamplePick a stratified sample, by city, of 20 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number.
| 1. __________ | 6. __________ | 11. __________ | 16. __________ |
| 2. __________ | 7. __________ | 12. __________ | 17. __________ |
| 3. __________ | 8. __________ | 13. __________ | 18. __________ |
| 4. __________ | 9. __________ | 14. __________ | 19. __________ |
| 5. __________ | 10. __________ | 15. __________ | 20. __________ |
A Stratified SamplePick a stratified sample, by entree cost, of 21 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number.
| 1. __________ | 6. __________ | 11. __________ | 16. __________ |
| 2. __________ | 7. __________ | 12. __________ | 17. __________ |
| 3. __________ | 8. __________ | 13. __________ | 18. __________ |
| 4. __________ | 9. __________ | 14. __________ | 19. __________ |
| 5. __________ | 10. __________ | 15. __________ | 20. __________ |
| 21. __________ |
A Cluster SamplePick a cluster sample of restaurants from two cities. The number of restaurants will vary.
| 1. ________ | 6. ________ | 11. ________ | 16. ________ | 21. ________ |
| 2. ________ | 7. ________ | 12. ________ | 17. ________ | 22. ________ |
| 3. ________ | 8. ________ | 13. ________ | 18. ________ | 23. ________ |
| 4. ________ | 9. ________ | 14. ________ | 19. ________ | 24. ________ |
| 5. ________ | 10. ________ | 15. ________ | 20. ________ | 25. ________ |
The mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text.
Data are individual items of information that come from a population or sample. Data may be classified as qualitative, quantitative continuous, or quantitative discrete.
Because it is not practical to measure the entire population in a study, researchers use samples to represent the population. A random sample is a representative group from the population chosen by using a method that gives each individual in the population an equal chance of being included in the sample. Random sampling methods include simple random sampling, stratified sampling, cluster sampling, and systematic sampling. Convenience sampling is a nonrandom method of choosing a sample that often produces biased data.
Samples that contain different individuals result in different data. This is true even when the samples are well-chosen and representative of the population. When properly selected, larger samples model the population more closely than smaller samples. There are many different potential problems that can affect the reliability of a sample. Statistical data needs to be critically analyzed, not simply accepted.
Some calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth.
In addition to rounding your answers, you can measure your data using the following four levels of measurement.
When organizing data, it is important to know how many times a value appears. How many statistics students study five hours or more for an exam? What percent of families on our block own two pets? Frequency, relative frequency, and cumulative relative frequency are measures that answer questions like these.
A poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. Participants in the control group receive a placebo treatment that looks exactly like the active treatments but cannot influence the response variable. To preserve the integrity of the placebo, both researchers and subjects may be blinded. When a study is designed properly, the only difference between treatment groups is the one imposed by the researcher. Therefore, when groups respond differently to different treatments, the difference must be due to the influence of the explanatory variable.
“An ethics problem arises when you are considering an action that benefits you or some cause you support, hurts or reduces benefits to others, and violates some rule.”4 Ethical violations in statistics are not always easy to spot. Professional associations and federal agencies post guidelines for proper conduct. It is important that you learn basic statistical procedures so that you can recognize proper data analysis.
Use the following information to answer the next five exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average (mean) length of time in months patients live once they start the treatment. Two researchers each follow a different set of 40 patients with AIDS from the start of treatment until their deaths. The following data (in months) are collected.
Researcher A:3 4 11 15 16 17 22 44 37 16 14 24 25 15 26 27 33 29 35 44 13 21 22 10 12 8 40 32 26 27 31 34 29 17 8 24 18 47 33 34
Researcher B:3 14 11 5 16 17 28 41 31 18 14 14 26 25 21 22 31 2 35 44 23 21 21 16 12 18 41 22 16 25 33 34 29 13 18 24 23 42 33 29
Determine what the key terms refer to in the example for Researcher A.
population
sample
parameter
statistic
variable
“Number of times per week” is what type of data?
a. qualitative b. quantitative discrete c. quantitative continuous
Use the following information to answer the next four exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Antonio, Texas. The first house in the neighborhood around the park was selected randomly, and then the resident of every eighth house in the neighborhood around the park was interviewed.
The sampling method was
a. simple random b. systematic c. stratified d. cluster
“Duration (amount of time)” is what type of data?
a. qualitative b. quantitative discrete c. quantitative continuous
The colors of the houses around the park are what kind of data?
a. qualitative b. quantitative discrete c. quantitative continuous
The population is ______________________
Table 1.26 contains the total number of deaths worldwide as a result of earthquakes from 2000 to 2012.
| Year | Total Number of Deaths |
|---|---|
| 2000 | 231 |
| 2001 | 21,357 |
| 2002 | 11,685 |
| 2003 | 33,819 |
| 2004 | 228,802 |
| 2005 | 88,003 |
| 2006 | 6,605 |
| 2007 | 712 |
| 2008 | 88,011 |
| 2009 | 1,790 |
| 2010 | 320,120 |
| 2011 | 21,953 |
| 2012 | 768 |
| Total | 823,856 |
Use Table 1.26 to answer the following questions.
For the following four exercises, determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
A group of test subjects is divided into twelve groups; then four of the groups are chosen at random.
A market researcher polls every tenth person who walks into a store.
The first 50 people who walk into a sporting event are polled on their television preferences.
A computer generates 100 random numbers, and 100 people whose names correspond with the numbers on the list are chosen.
Researcher A: 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B: 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Complete the tables using the data provided:
| Survival Length (in months) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0.5–6.5 | |||
| 6.5–12.5 | |||
| 12.5–18.5 | |||
| 18.5–24.5 | |||
| 24.5–30.5 | |||
| 30.5–36.5 | |||
| 36.5–42.5 | |||
| 42.5–48.5 |
| Survival Length (in months) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0.5–6.5 | |||
| 6.5–12.5 | |||
| 12.5–18.5 | |||
| 18.5–24.5 | |||
| 24.5–30.5 | |||
| 30.5–36.5 | |||
| 36.5-45.5 |
Determine what the key term data refers to in the above example for Researcher A.
List two reasons why the data may differ.
Can you tell if one researcher is correct and the other one is incorrect? Why?
Would you expect the data to be identical? Why or why not?
How might the researchers gather random data?
Suppose that the first researcher conducted his survey by randomly choosing one state in the nation and then randomly picking 40 patients from that state. What sampling method would that researcher have used?
Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What sampling method would that researcher have used? What concerns would you have about this data set, based upon the data collection method?
Use the following data to answer the next five exercises: Two researchers are gathering data on hours of video games played by school-aged children and young adults. They each randomly sample different groups of 150 students from the same school. They collect the following data.
| Hours Played per Week | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0–2 | 26 | 0.17 | 0.17 |
| 2–4 | 30 | 0.20 | 0.37 |
| 4–6 | 49 | 0.33 | 0.70 |
| 6–8 | 25 | 0.17 | 0.87 |
| 8–10 | 12 | 0.08 | 0.95 |
| 10–12 | 8 | 0.05 | 1 |
| Hours Played per Week | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0–2 | 48 | 0.32 | 0.32 |
| 2–4 | 51 | 0.34 | 0.66 |
| 4–6 | 24 | 0.16 | 0.82 |
| 6–8 | 12 | 0.08 | 0.90 |
| 8–10 | 11 | 0.07 | 0.97 |
| 10–12 | 4 | 0.03 | 1 |
Give a reason why the data may differ.
Would the sample size be large enough if the population is the students in the school?
Would the sample size be large enough if the population is school-aged children and young adults in the United States?
Researcher A concludes that most students play video games between four and six hours each week. Researcher B concludes that most students play video games between two and four hours each week. Who is correct?
As part of a way to reward students for participating in the survey, the researchers gave each student a gift card to a video game store. Would this affect the data if students knew about the award before the study?
Use the following data to answer the next five exercises: A pair of studies was performed to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The studies observed 200 stroke patients recovering over a period of several weeks. The first study collected the data in Table 1.31. The second study collected the data in Table 1.32.
| Group | Showed improvement | No improvement | Deterioration |
|---|---|---|---|
| Used program | 142 | 43 | 15 |
| Did not use program | 72 | 110 | 18 |
| Group | Showed improvement | No improvement | Deterioration |
|---|---|---|---|
| Used program | 105 | 74 | 19 |
| Did not use program | 89 | 99 | 12 |
Given what you know, which study is correct?
The first study was performed by the company that designed the software program. The second study was performed by the American Medical Association. Which study is more reliable?
Both groups that performed the study concluded that the software works. Is this accurate?
The company takes the two studies as proof that their software causes mental improvement in stroke patients. Is this a fair statement?
Patients who used the software were also a part of an exercise program whereas patients who did not use the software were not. Does this change the validity of the conclusions from Exercise 1.31?
Is a sample size of 1,000 a reliable measure for a population of 5,000?
Is a sample of 500 volunteers a reliable measure for a population of 2,500?
A question on a survey reads: "Do you prefer the delicious taste of Brand X or the taste of Brand Y?" Is this a fair question?
Is a sample size of two representative of a population of five?
Is it possible for two experiments to be well run with similar sample sizes to get different data?
What type of measure scale is being used? Nominal, ordinal, interval or ratio.
Design an experiment. Identify the explanatory and response variables. Describe the population being studied and the experimental units. Explain the treatments that will be used and how they will be assigned to the experimental units. Describe how blinding and placebos may be used to counter the power of suggestion.
Discuss potential violations of the rule requiring informed consent.
For each of the following eight exercises, identify: a. the population, b. the sample, c. the parameter, d. the statistic, e. the variable, and f. the data. Give examples where appropriate.
A fitness center is interested in the mean amount of time a client exercises in the center each week.
Ski resorts are interested in the mean age that children take their first ski and snowboard lessons. They need this information to plan their ski classes optimally.
A cardiologist is interested in the mean recovery period of her patients who have had heart attacks.
Insurance companies are interested in the mean health costs each year of their clients, so that they can determine the costs of health insurance.
A politician is interested in the proportion of voters in his district who think he is doing a good job.
A marriage counselor is interested in the proportion of clients she counsels who stay married.
Political pollsters may be interested in the proportion of people who will vote for a particular cause.
A marketing company is interested in the proportion of people who will buy a particular product.
What is the population she is interested in?
Consider the following:
= number of days a Lake Tahoe Community College math student is absent
In this case, X is an example of a:
The instructor’s sample produces a mean number of days absent of 3.5 days. This value is an example of a:
For the following exercises, identify the type of data that would be used to describe a response (quantitative discrete, quantitative continuous, or qualitative), and give an example of the data.
number of tickets sold to a concert
percent of body fat
favorite baseball team
time in line to buy groceries
number of students enrolled at Evergreen Valley College
most-watched television show
brand of toothpaste
distance to the closest movie theatre
age of executives in Fortune 500 companies
number of competing computer spreadsheet software packages
Use the following information to answer the next two exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of resident use of a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every 8th house in the neighborhood around the park was interviewed.
“Number of times per week” is what type of data?
“Duration (amount of time)” is what type of data?
Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. Suppose an airline conducts a survey. Over Thanksgiving weekend, it surveys six flights from Boston to Salt Lake City to determine the number of babies on the flights. It determines the amount of safety equipment needed by the result of that study.
Suppose you want to determine the mean number of students per statistics class in your state. Describe a possible sampling method in three to five complete sentences. Make the description detailed.
Suppose you want to determine the mean number of cans of soda drunk each month by students in their twenties at your school. Describe a possible sampling method in three to five complete sentences. Make the description detailed.
List some practical difficulties involved in getting accurate results from a telephone survey.
List some practical difficulties involved in getting accurate results from a mailed survey.
With your classmates, brainstorm some ways you could overcome these problems if you needed to conduct a phone or mail survey.
The instructor takes her sample by gathering data on five randomly selected students from each Lake Tahoe Community College math class. The type of sampling she used is
A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every eighth house in the neighborhood around the park was interviewed. The sampling method was:
Name the sampling method used in each of the following situations:
A “random survey” was conducted of 3,274 people of the “microprocessor generation” (people born since 1971, the year the microprocessor was invented). It was reported that 48% of those individuals surveyed stated that if they had $2,000 to spend, they would use it for computer equipment. Also, 66% of those surveyed considered themselves relatively savvy computer users.
The Gallup-Healthways Well-Being Index is a survey that follows trends of U.S. residents on a regular basis. There are six areas of health and wellness covered in the survey: Life Evaluation, Emotional Health, Physical Health, Healthy Behavior, Work Environment, and Basic Access. Some of the questions used to measure the Index are listed below.
Identify the type of data obtained from each question used in this survey: qualitative, quantitative discrete, or quantitative continuous.
In advance of the 1936 Presidential Election, a magazine titled Literary Digest released the results of an opinion poll predicting that the republican candidate Alf Landon would win by a large margin. The magazine sent post cards to approximately 10,000,000 prospective voters. These prospective voters were selected from the subscription list of the magazine, from automobile registration lists, from phone lists, and from club membership lists. Approximately 2,300,000 people returned the postcards.
Crime-related and demographic statistics for 47 US states in 1960 were collected from government agencies, including the FBI's Uniform Crime Report. One analysis of this data found a strong connection between education and crime indicating that higher levels of education in a community correspond to higher crime rates.
Which of the potential problems with samples discussed in Data, Sampling, and Variation in Data and Sampling could explain this connection?
YouPolls is a website that allows anyone to create and respond to polls. One question posted April 15 asks:
“Do you feel happy paying your taxes when members of the Obama administration are allowed to ignore their tax liabilities?”1
As of April 25, 11 people responded to this question. Each participant answered “NO!”
Which of the potential problems with samples discussed in this module could explain this connection?
A scholarly article about response rates begins with the following quote:
“Declining contact and cooperation rates in random digit dial (RDD) national telephone surveys raise serious concerns about the validity of estimates drawn from such research.”2
The Pew Research Center for People and the Press admits:
“The percentage of people we interview – out of all we try to interview – has been declining over the past decade or more.”3
Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below:
| # of Courses | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 1 | 30 | 0.6 | |
| 2 | 15 | ||
| 3 |
Sixty adults with gum disease were asked the number of times per week they used to floss before their diagnosis. The (incomplete) results are shown in Table 1.34.
| # Flossing per Week | Frequency | Relative Frequency | Cumulative Relative Freq. |
|---|---|---|---|
| 0 | 27 | 0.4500 | |
| 1 | 18 | ||
| 3 | 0.9333 | ||
| 6 | 3 | 0.0500 | |
| 7 | 1 | 0.0167 |
Nineteen immigrants to the U.S were asked how many years, to the nearest year, they have lived in the U.S. The data are as follows: 2 5 7 2 2 10 20 15 0 7 0 20 5 12 15 12 4 5 10 .
Table 1.35 was produced.
| Data | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 2 | 0.1053 | |
| 2 | 3 | 0.2632 | |
| 4 | 1 | 0.3158 | |
| 5 | 3 | 0.4737 | |
| 7 | 2 | 0.5789 | |
| 10 | 2 | 0.6842 | |
| 12 | 2 | 0.7895 | |
| 15 | 1 | 0.8421 | |
| 20 | 1 | 1.0000 |
How much time does it take to travel to work? Table 1.36 shows the mean commute time by state for workers at least 16 years old who are not working at home. Find the mean travel time, and round off the answer properly.
| 24.0 | 24.3 | 25.9 | 18.9 | 27.5 | 17.9 | 21.8 | 20.9 | 16.7 | 27.3 |
| 18.2 | 24.7 | 20.0 | 22.6 | 23.9 | 18.0 | 31.4 | 22.3 | 24.0 | 25.5 |
| 24.7 | 24.6 | 28.1 | 24.9 | 22.6 | 23.6 | 23.4 | 25.7 | 24.8 | 25.5 |
| 21.2 | 25.7 | 23.1 | 23.0 | 23.9 | 26.0 | 16.3 | 23.1 | 21.4 | 21.5 |
| 27.0 | 27.0 | 18.6 | 31.7 | 23.3 | 30.1 | 22.9 | 23.3 | 21.7 | 18.6 |
Forbes magazine published data on the best small firms in 2012. These were firms which had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. Table 1.37 shows the ages of the chief executive officers for the first 60 ranked firms.
| Age | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 40–44 | 3 | ||
| 45–49 | 11 | ||
| 50–54 | 13 | ||
| 55–59 | 16 | ||
| 60–64 | 10 | ||
| 65–69 | 6 | ||
| 70–74 | 1 |
Use the following information to answer the next two exercises: Table 1.38 contains data on hurricanes that have made direct hits on the U.S. Between 1851 and 2004. A hurricane is given a strength category rating based on the minimum wind speed generated by the storm.
| Category | Number of Direct Hits | Relative Frequency | Cumulative Frequency |
|---|---|---|---|
| Total = 273 | |||
| 1 | 109 | 0.3993 | 0.3993 |
| 2 | 72 | 0.2637 | 0.6630 |
| 3 | 71 | 0.2601 | |
| 4 | 18 | 0.9890 | |
| 5 | 3 | 0.0110 | 1.0000 |
What is the relative frequency of direct hits that were category 4 hurricanes?
What is the relative frequency of direct hits that were AT MOST a category 3 storm?
How does sleep deprivation affect your ability to drive? A recent study measured the effects on 19 professional drivers. Each driver participated in two experimental sessions: one after normal sleep and one after 27 hours of total sleep deprivation. The treatments were assigned in random order. In each session, performance was measured on a variety of tasks including a driving simulation.
Use key terms from this module to describe the design of this experiment.
An advertisement for Acme Investments displays the two graphs in Figure 1.14 to show the value of Acme’s product in comparison with the Other Guy’s product. Describe the potentially misleading visual effect of these comparison graphs. How can this be corrected?
The graph in Figure 1.15 shows the number of complaints for six different airlines as reported to the US Department of Transportation in February 2013. Alaska, Pinnacle, and Airtran Airlines have far fewer complaints reported than American, Delta, and United. Can we conclude that American, Delta, and United are the worst airline carriers since they have the most complaints?
Seven hundred and seventy-one distance learning students at Long Beach City College responded to surveys in the 2010-11 academic year. Highlights of the summary report are listed in Table 1.39.
| Have computer at home | 96% |
| Unable to come to campus for classes | 65% |
| Age 41 or over | 24% |
| Would like LBCC to offer more DL courses | 95% |
| Took DL classes due to a disability | 17% |
| Live at least 16 miles from campus | 13% |
| Took DL courses to fulfill transfer requirements | 71% |
Several online textbook retailers advertise that they have lower prices than on-campus bookstores. However, an important factor is whether the Internet retailers actually have the textbooks that students need in stock. Students need to be able to get textbooks promptly at the beginning of the college term. If the book is not available, then a student would not be able to get the textbook at all, or might get a delayed delivery if the book is back ordered.
Write an analysis of his study that addresses the following issues: Is his sample representative of the population of all college textbooks? Explain why or why not. Describe some possible sources of bias in this study, and how it might affect the results of the study. Give some suggestions about what could be done to improve the study.
The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/CrashTestDummies.html (accessed May 1, 2013).
Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/default.asp (accessed May 1, 2013).
Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/methodology.asp (accessed May 1, 2013).
Gallup-Healthways Well-Being Index. http://www.gallup.com/poll/146822/gallup-healthways-index-questions.aspx (accessed May 1, 2013).
Data from http://www.bookofodds.com/Relationships-Society/Articles/A0374-How-George-Gallup-Picked-the-President
Dominic Lusinchi, “’President’ Landon and the 1936 Literary Digest Poll: Were Automobile and Telephone Owners to Blame?” Social Science History 36, no. 1: 23-54 (2012), http://ssh.dukejournals.org/content/36/1/23.abstract (accessed May 1, 2013).
“The Literary Digest Poll,” Virtual Laboratories in Probability and Statistics http://www.math.uah.edu/stat/data/LiteraryDigest.html (accessed May 1, 2013).
“Gallup Presidential Election Trial-Heat Trends, 1936–2008,” Gallup Politics http://www.gallup.com/poll/110548/gallup-presidential-election-trialheat-trends-19362004.aspx#4 (accessed May 1, 2013).
The Data and Story Library, http://lib.stat.cmu.edu/DASL/Datafiles/USCrime.html (accessed May 1, 2013).
LBCC Distance Learning (DL) program data in 2010-2011, http://de.lbcc.edu/reports/2010-11/future/highlights.html#focus (accessed May 1, 2013).
Data from San Jose Mercury News
“State & County QuickFacts,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/download_data.html (accessed May 1, 2013).
“State & County QuickFacts: Quick, easy access to facts about people, business, and geography,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/index.html (accessed May 1, 2013).
“Table 5: Direct hits by mainland United States Hurricanes (1851-2004),” National Hurricane Center, http://www.nhc.noaa.gov/gifs/table5.gif (accessed May 1, 2013).
“Levels of Measurement,” http://infinity.cos.edu/faculty/woodbury/stats/tutorial/Data_Levels.htm (accessed May 1, 2013).
Courtney Taylor, “Levels of Measurement,” about.com, http://statistics.about.com/od/HelpandTutorials/a/Levels-Of-Measurement.htm (accessed May 1, 2013).
David Lane. “Levels of Measurement,” Connexions, http://cnx.org/content/m10809/latest/ (accessed May 1, 2013).
“Vitamin E and Health,” Nutrition Source, Harvard School of Public Health, http://www.hsph.harvard.edu/nutritionsource/vitamin-e/ (accessed May 1, 2013).
Stan Reents. “Don’t Underestimate the Power of Suggestion,” athleteinme.com, http://www.athleteinme.com/ArticleView.aspx?id=1053 (accessed May 1, 2013).
Ankita Mehta. “Daily Dose of Aspiring Helps Reduce Heart Attacks: Study,” International Business Times, July 21, 2011. Also available online at http://www.ibtimes.com/daily-dose-aspirin-helps-reduce-heart-attacks-study-300443 (accessed May 1, 2013).
The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/ScentsandLearning.html (accessed May 1, 2013).
M.L. Jacskon et al., “Cognitive Components of Simulated Driving Performance: Sleep Loss effect and Predictors,” Accident Analysis and Prevention Journal, Jan no. 50 (2013), http://www.ncbi.nlm.nih.gov/pubmed/22721550 (accessed May 1, 2013).
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Data from www.businessweek.com (accessed May 1, 2013).
Data from www.forbes.com (accessed May 1, 2013).
“America’s Best Small Companies,” http://www.forbes.com/best-small-companies/list/ (accessed May 1, 2013).
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AIDS patients.
The average length of time (in months) AIDS patients live after treatment.
X = the length of time (in months) AIDS patients live after treatment
b
a
systematic
simple random
values for X, such as 3, 4, 11, and so on
No, we do not have enough information to make such a claim.
Take a simple random sample from each group. One way is by assigning a number to each patient and using a random number generator to randomly select patients.
This would be convenience sampling and is not random.
Yes, the sample size of 150 would be large enough to reflect a population of one school.
Even though the specific data support each researcher’s conclusions, the different results suggest that more data need to be collected before the researchers can reach a conclusion.
There is not enough information given to judge if either one is correct or incorrect.
The software program seems to work because the second study shows that more patients improve while using the software than not. Even though the difference is not as large as that in the first study, the results from the second study are likely more reliable and still show improvement.
Yes, because we cannot tell if the improvement was due to the software or the exercise; the data is confounded, and a reliable conclusion cannot be drawn. New studies should be performed.
No, even though the sample is large enough, the fact that the sample consists of volunteers makes it a self-selected sample, which is not reliable.
No, even though the sample is a large portion of the population, two responses are not enough to justify any conclusions. Because the population is so small, it would be better to include everyone in the population to get the most accurate data.
a
quantitative discrete, 150
qualitative, Oakland A’s
quantitative discrete, 11,234 students
qualitative, Crest
quantitative continuous, 47.3 years
b
Answers will vary. Sample Answer: You could use a systematic sampling method. Stop the tenth person as they leave one of the buildings on campus at 9:50 in the morning. Then stop the tenth person as they leave a different building on campus at 1:50 in the afternoon.
Answers will vary. Sample Answer: Many people will not respond to mail surveys. If they do respond to the surveys, you can’t be sure who is responding. In addition, mailing lists can be incomplete.
b
convenience cluster stratified systematic simple random
Causality: The fact that two variables are related does not guarantee that one variable is influencing the other. We cannot assume that crime rate impacts education level or that education level impacts crime rate.
Confounding: There are many factors that define a community other than education level and crime rate. Communities with high crime rates and high education levels may have other lurking variables that distinguish them from communities with lower crime rates and lower education levels. Because we cannot isolate these variables of interest, we cannot draw valid conclusions about the connection between education and crime. Possible lurking variables include police expenditures, unemployment levels, region, average age, and size.
| # Flossing per Week | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 27 | 0.4500 | 0.4500 |
| 1 | 18 | 0.3000 | 0.7500 |
| 3 | 11 | 0.1833 | 0.9333 |
| 6 | 3 | 0.0500 | 0.9833 |
| 7 | 1 | 0.0167 | 1 |
The sum of the travel times is 1,173.1. Divide the sum by 50 to calculate the mean value: 23.462. Because each state’s travel time was measured to the nearest tenth, round this calculation to the nearest hundredth: 23.46.
b
Explanatory variable: amount of sleep
You cannot assume that the numbers of complaints reflect the quality of the airlines. The airlines shown with the greatest number of complaints are the ones with the most passengers. You must consider the appropriateness of methods for presenting data; in this case displaying totals is misleading.
Answers will vary. Sample answer: The sample is not representative of the population of all college textbooks. Two reasons why it is not representative are that he only sampled seven subjects and he only investigated one textbook in each subject. There are several possible sources of bias in the study. The seven subjects that he investigated are all in mathematics and the sciences; there are many subjects in the humanities, social sciences, and other subject areas, (for example: literature, art, history, psychology, sociology, business) that he did not investigate at all. It may be that different subject areas exhibit different patterns of textbook availability, but his sample would not detect such results.
He also looked only at the most popular textbook in each of the subjects he investigated. The availability of the most popular textbooks may differ from the availability of other textbooks in one of two ways:
In reality, many college students do not use the most popular textbook in their subject, and this study gives no useful information about the situation for those less popular textbooks.
He could improve this study by:
By the end of this chapter, the student should be able to:
Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data.
In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called "Descriptive Statistics." You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs.
A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied.
Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. Our emphasis will be on histograms and box plots.
This book contains instructions for constructing a histogram and a box plot for the TI-83+ and TI-84 calculators. The Texas Instruments (TI) website provides additional instructions for using these calculators.
One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.
For Susan Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest):
| Stem | Leaf |
|---|---|
| 3 | 3 |
| 4 | 2 9 9 |
| 5 | 3 5 5 |
| 6 | 1 3 7 8 8 9 9 |
| 7 | 2 3 4 8 |
| 8 | 0 3 8 8 8 |
| 9 | 0 2 4 4 4 4 6 |
| 10 | 0 |
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% were in the 90s or 100, a fairly high number of As.
For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):
| Stem | Leaf |
|---|---|
| 3 | 2 2 3 4 8 |
| 4 | 0 2 2 3 4 6 7 7 8 8 8 9 |
| 5 | 0 0 1 2 2 2 3 4 6 7 7 |
| 6 | 0 1 |
The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.
The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data:
Do the data seem to have any concentration of values?
The leaves are to the right of the decimal.
The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.
| Stem | Leaf |
|---|---|
| 1 | 1 5 |
| 2 | 3 5 7 |
| 3 | 2 3 3 5 8 |
| 4 | 0 2 5 5 7 8 |
| 5 | 5 6 |
| 6 | 5 7 |
| 7 | |
| 8 | |
| 9 | |
| 10 | |
| 11 | |
| 12 | 3 |
The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers:
0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0
| Stem | Leaf |
|---|---|
| 0 | 5 7 |
| 1 | 1 2 2 3 3 5 5 7 7 8 9 |
| 2 | 0 2 5 6 8 8 8 |
| 3 | 5 8 |
| 4 | 4 8 9 |
| 5 | 2 5 7 8 |
| 6 | |
| 7 | |
| 8 | 0 |
The value 8.0 may be an outlier. Values appear to concentrate at one and two miles.
A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. Table 2.6 and Table 2.7 show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.
| Ages at Inauguration | Ages at Death | |
|---|---|---|
| 9 9 8 7 7 7 6 3 2 | 4 | 6 9 |
| 8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 1 1 1 1 1 0 | 5 | 3 6 6 7 7 8 |
| 9 5 4 4 2 1 1 1 0 | 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |
| 7 | 0 0 1 1 1 4 7 8 8 9 | |
| 8 | 0 1 3 5 8 | |
| 9 | 0 0 3 3 |
| President | Age | President | Age | President | Age |
|---|---|---|---|---|---|
| Washington | 57 | Lincoln | 52 | Hoover | 54 |
| J. Adams | 61 | A. Johnson | 56 | F. Roosevelt | 51 |
| Jefferson | 57 | Grant | 46 | Truman | 60 |
| Madison | 57 | Hayes | 54 | Eisenhower | 62 |
| Monroe | 58 | Garfield | 49 | Kennedy | 43 |
| J. Q. Adams | 57 | Arthur | 51 | L. Johnson | 55 |
| Jackson | 61 | Cleveland | 47 | Nixon | 56 |
| Van Buren | 54 | B. Harrison | 55 | Ford | 61 |
| W. H. Harrison | 68 | Cleveland | 55 | Carter | 52 |
| Tyler | 51 | McKinley | 54 | Reagan | 69 |
| Polk | 49 | T. Roosevelt | 42 | G.H.W. Bush | 64 |
| Taylor | 64 | Taft | 51 | Clinton | 47 |
| Fillmore | 50 | Wilson | 56 | G. W. Bush | 54 |
| Pierce | 48 | Harding | 55 | Obama | 47 |
| Buchanan | 65 | Coolidge | 51 |
| President | Age | President | Age | President | Age |
|---|---|---|---|---|---|
| Washington | 67 | Lincoln | 56 | Hoover | 90 |
| J. Adams | 90 | A. Johnson | 66 | F. Roosevelt | 63 |
| Jefferson | 83 | Grant | 63 | Truman | 88 |
| Madison | 85 | Hayes | 70 | Eisenhower | 78 |
| Monroe | 73 | Garfield | 49 | Kennedy | 46 |
| J. Q. Adams | 80 | Arthur | 56 | L. Johnson | 64 |
| Jackson | 78 | Cleveland | 71 | Nixon | 81 |
| Van Buren | 79 | B. Harrison | 67 | Ford | 93 |
| W. H. Harrison | 68 | Cleveland | 71 | Reagan | 93 |
| Tyler | 71 | McKinley | 58 | ||
| Polk | 53 | T. Roosevelt | 60 | ||
| Taylor | 65 | Taft | 72 | ||
| Fillmore | 74 | Wilson | 67 | ||
| Pierce | 64 | Harding | 57 | ||
| Buchanan | 77 | Coolidge | 60 |
The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.
| Losses | Wins | Year | Losses | Wins | Year |
|---|---|---|---|---|---|
| 34 | 48 | 1968–1969 | 41 | 41 | 1989–1990 |
| 34 | 48 | 1969–1970 | 39 | 43 | 1990–1991 |
| 46 | 36 | 1970–1971 | 44 | 38 | 1991–1992 |
| 46 | 36 | 1971–1972 | 39 | 43 | 1992–1993 |
| 36 | 46 | 1972–1973 | 25 | 57 | 1993–1994 |
| 47 | 35 | 1973–1974 | 40 | 42 | 1994–1995 |
| 51 | 31 | 1974–1975 | 36 | 46 | 1995–1996 |
| 53 | 29 | 1975–1976 | 26 | 56 | 1996–1997 |
| 51 | 31 | 1976–1977 | 32 | 50 | 1997–1998 |
| 41 | 41 | 1977–1978 | 19 | 31 | 1998–1999 |
| 36 | 46 | 1978–1979 | 54 | 28 | 1999–2000 |
| 32 | 50 | 1979–1980 | 57 | 25 | 2000–2001 |
| 51 | 31 | 1980–1981 | 49 | 33 | 2001–2002 |
| 40 | 42 | 1981–1982 | 47 | 35 | 2002–2003 |
| 39 | 43 | 1982–1983 | 54 | 28 | 2003–2004 |
| 42 | 40 | 1983–1984 | 69 | 13 | 2004–2005 |
| 48 | 34 | 1984–1985 | 56 | 26 | 2005–2006 |
| 32 | 50 | 1985–1986 | 52 | 30 | 2006–2007 |
| 25 | 57 | 1986–1987 | 45 | 37 | 2007–2008 |
| 32 | 50 | 1987–1988 | 35 | 47 | 2008–2009 |
| 30 | 52 | 1988–1989 | 29 | 53 | 2009–2010 |
| Atlanta Hawks Wins and Losses | ||
|---|---|---|
| Number of Wins | Number of Losses | |
| 3 | 1 | 9 |
| 9 8 8 6 5 | 2 | 5 5 9 |
| 8 7 6 6 5 5 4 3 1 1 1 1 0 | 3 | 0 2 2 2 2 4 4 5 6 6 6 9 9 9 |
| 8 8 7 6 6 6 3 3 3 2 2 1 1 0 | 4 | 0 0 1 1 2 4 5 6 6 7 7 8 9 |
| 7 7 6 3 2 0 0 0 0 | 5 | 1 1 1 2 3 4 4 6 7 |
| 6 | 9 | |
Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in Example 2.3, the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments.
In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in Table 2.10 and in Figure 2.2.
| Number of times teenager is reminded | Frequency |
|---|---|
| 0 | 2 |
| 1 | 5 |
| 2 | 8 |
| 3 | 14 |
| 4 | 7 |
| 5 | 4 |
In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in Table 2.11. Construct a line graph.
| Number of times in shop | Frequency |
|---|---|
| 0 | 7 |
| 1 | 10 |
| 2 | 14 |
| 3 | 9 |
Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in Example 2.4 has age groups represented on the x-axis and proportions on the y-axis.
By the end of 2011, Facebook had over 146 million users in the United States. Table 2.12 shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data.
| Age groups | Number of Facebook users | Proportion (%) of Facebook users |
|---|---|---|
| 13–25 | 65,082,280 | 45% |
| 26–44 | 53,300,200 | 36% |
| 45–64 | 27,885,100 | 19% |
The population in Park City is made up of children, working-age adults, and retirees. Table 2.13 shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions.
| Age groups | Number of people | Proportion of population |
|---|---|---|
| Children | 67,059 | 19% |
| Working-age adults | 152,198 | 43% |
| Retirees | 131,662 | 38% |
The columns in Table 2.14 contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis.
| Race/Ethnicity | AP Examinee Population | Overall Student Population |
|---|---|---|
| 1 = Asian, Asian American or Pacific Islander | 10.3% | 5.7% |
| 2 = Black or African American | 9.0% | 14.7% |
| 3 = Hispanic or Latino | 17.0% | 17.6% |
| 4 = American Indian or Alaska Native | 0.6% | 1.1% |
| 5 = White | 57.1% | 59.2% |
| 6 = Not reported/other | 6.0% | 1.7% |
Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.
| District | Registered voter population | Overall city population |
|---|---|---|
| 1 | 15.5% | 19.4% |
| 2 | 12.2% | 15.6% |
| 3 | 9.8% | 9.0% |
| 4 | 17.4% | 18.5% |
| 5 | 22.8% | 20.7% |
| 6 | 22.3% | 16.8% |
For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.
The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample.(Remember, frequency is defined as the number of times an answer occurs.) If:
then:
For example, if three students in Mr. Ahab's English class of 40 students received from 90% to 100%, then, f = 3, n = 40, and RF = = = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures.
To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.
The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured.
The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.
60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.
The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.
Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars.
We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.
The boundaries are:
The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.
The following histogram displays the heights on the x-axis and relative frequency on the y-axis.
The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
Smallest value: 9
Largest value: 14
Convenient starting value: 9 – 0.05 = 8.95
Convenient ending value: 14 + 0.05 = 14.05
The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one.
The following data are the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted.
Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.
Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.
Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .
Calculate the number of bars as follows:
where 1 is the width of a bar. Therefore, bars = 6.
The following histogram displays the number of books on the x-axis and the frequency on the y-axis.
Go to Table 13.45. There are calculator instructions for entering data and for creating a customized histogram. Create the histogram for Example 2.7.
The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
Fill in the blanks for the following sentence. Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to _____, the 2 in the middle of the interval from _____ to _____, and the 3 in the middle of the interval from _____ to _____.
1.5
Using this data set, construct a histogram.
| Number of Hours My Classmates Spent Playing Video Games on Weekends | ||||
|---|---|---|---|---|
| 9.95 | 10 | 2.25 | 16.75 | 0 |
| 19.5 | 22.5 | 7.5 | 15 | 12.75 |
| 5.5 | 11 | 10 | 20.75 | 17.5 |
| 23 | 21.9 | 24 | 23.75 | 18 |
| 20 | 15 | 22.9 | 18.8 | 20.5 |
Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram.
The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram.
22351526 40281820 25343942 24221927 22344020 38and 28
Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals.
Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons.
To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x-axis and y-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them.
A frequency polygon was constructed from the frequency table below.
| Frequency Distribution for Calculus Final Test Scores | |||
|---|---|---|---|
| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |
| 49.5 | 59.5 | 5 | 5 |
| 59.5 | 69.5 | 10 | 15 |
| 69.5 | 79.5 | 30 | 45 |
| 79.5 | 89.5 | 40 | 85 |
| 89.5 | 99.5 | 15 | 100 |
The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in Table 2.18.
| Age at Inauguration | Frequency |
|---|---|
| 41.5–46.5 | 4 |
| 46.5–51.5 | 11 |
| 51.5–56.5 | 14 |
| 56.5–61.5 | 9 |
| 61.5–66.5 | 4 |
| 66.5–71.5 | 2 |
The first label on the x-axis is 39. This represents an interval extending from 36.5 to 41.5. Since there are no ages less than 41.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 44 represents the next interval, or the first “real” interval from the table, and contains four scores. This reasoning is followed for each of the remaining intervals with the point 74 representing the interval from 71.5 to 76.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets.
We will construct an overlay frequency polygon comparing the scores from Example 2.9 with the students’ final numeric grade.
| Frequency Distribution for Calculus Final Test Scores | |||
|---|---|---|---|
| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |
| 49.5 | 59.5 | 5 | 5 |
| 59.5 | 69.5 | 10 | 15 |
| 69.5 | 79.5 | 30 | 45 |
| 79.5 | 89.5 | 40 | 85 |
| 89.5 | 99.5 | 15 | 100 |
| Frequency Distribution for Calculus Final Grades | |||
|---|---|---|---|
| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |
| 49.5 | 59.5 | 10 | 10 |
| 59.5 | 69.5 | 10 | 20 |
| 69.5 | 79.5 | 30 | 50 |
| 79.5 | 89.5 | 45 | 95 |
| 89.5 | 99.5 | 5 | 100 |
Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with this data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.
One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.
To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur.
The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only.
| Year | Jan | Feb | Mar | Apr | May | Jun | Jul |
|---|---|---|---|---|---|---|---|
| 2003 | 181.7 | 183.1 | 184.2 | 183.8 | 183.5 | 183.7 | 183.9 |
| 2004 | 185.2 | 186.2 | 187.4 | 188.0 | 189.1 | 189.7 | 189.4 |
| 2005 | 190.7 | 191.8 | 193.3 | 194.6 | 194.4 | 194.5 | 195.4 |
| 2006 | 198.3 | 198.7 | 199.8 | 201.5 | 202.5 | 202.9 | 203.5 |
| 2007 | 202.416 | 203.499 | 205.352 | 206.686 | 207.949 | 208.352 | 208.299 |
| 2008 | 211.080 | 211.693 | 213.528 | 214.823 | 216.632 | 218.815 | 219.964 |
| 2009 | 211.143 | 212.193 | 212.709 | 213.240 | 213.856 | 215.693 | 215.351 |
| 2010 | 216.687 | 216.741 | 217.631 | 218.009 | 218.178 | 217.965 | 218.011 |
| 2011 | 220.223 | 221.309 | 223.467 | 224.906 | 225.964 | 225.722 | 225.922 |
| 2012 | 226.665 | 227.663 | 229.392 | 230.085 | 229.815 | 229.478 | 229.104 |
| Year | Aug | Sep | Oct | Nov | Dec | Annual |
|---|---|---|---|---|---|---|
| 2003 | 184.6 | 185.2 | 185.0 | 184.5 | 184.3 | 184.0 |
| 2004 | 189.5 | 189.9 | 190.9 | 191.0 | 190.3 | 188.9 |
| 2005 | 196.4 | 198.8 | 199.2 | 197.6 | 196.8 | 195.3 |
| 2006 | 203.9 | 202.9 | 201.8 | 201.5 | 201.8 | 201.6 |
| 2007 | 207.917 | 208.490 | 208.936 | 210.177 | 210.036 | 207.342 |
| 2008 | 219.086 | 218.783 | 216.573 | 212.425 | 210.228 | 215.303 |
| 2009 | 215.834 | 215.969 | 216.177 | 216.330 | 215.949 | 214.537 |
| 2010 | 218.312 | 218.439 | 218.711 | 218.803 | 219.179 | 218.056 |
| 2011 | 226.545 | 226.889 | 226.421 | 226.230 | 225.672 | 224.939 |
| 2012 | 230.379 | 231.407 | 231.317 | 230.221 | 229.601 | 229.594 |
The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO2 emissions for the United States.
| CO2 Emissions | |||
|---|---|---|---|
| Ukraine | United Kingdom | United States | |
| 2003 | 352,259 | 540,640 | 5,681,664 |
| 2004 | 343,121 | 540,409 | 5,790,761 |
| 2005 | 339,029 | 541,990 | 5,826,394 |
| 2006 | 327,797 | 542,045 | 5,737,615 |
| 2007 | 328,357 | 528,631 | 5,828,697 |
| 2008 | 323,657 | 522,247 | 5,656,839 |
| 2009 | 272,176 | 474,579 | 5,299,563 |
Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot.
The common measures of location are quartiles and percentiles
Quartiles are special percentiles. The first quartile, Q1, is the same as the 25th percentile, and the third quartile, Q3, is the same as the 75th percentile. The median, M, is called both the second quartile and the 50th percentile.
To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.
Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75th percentile. That translates into a score of at least 1220.
Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.
The median is a number that measures the "center" of the data. You can think of the median as the "middle value," but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.
Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.
The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.
Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q1, is the middle value of the lower half of the data, and the third quartile, Q3, is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:
The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.
The number two, which is part of the data, is the first quartile. One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.
The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.
The third quartile, Q3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.
The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1).
IQR = Q3 – Q1
The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is less than (1.5)(IQR) below the first quartile or more than (1.5)(IQR) above the third quartile. Potential outliers always require further investigation.
A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.
For the following 13 real estate prices, calculate the IQR and determine if any prices are potential outliers. Prices are in dollars.
Order the data from smallest to largest.
M = 488,800
Q1 = = 308,750
Q3 = = 649,000
IQR = 649,000 – 308,750 = 340,250
(1.5)(IQR) = (1.5)(340,250) = 510,375
Q1 – (1.5)(IQR) = 308,750 – 510,375 = –201,625
Q3 + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375
No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier.
For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars.
$33,000 $64,500 $28,000 $54,000 $72,000 $68,500 $69,000 $42,000 $54,000 $120,000 $40,500
Order the data from smallest to largest.
$28,000 $33,000 $40,500 $42,000 $54,000 $54,000 $64,500 $68,500 $69,000 $72,000 $120,000
Median = $54,000
Q1 = $40,500
Q3 = $69,000
IQR = $69,000 – $40,500 = $28,500
(1.5)(IQR) = (1.5)($28,500) = $42,750
Q1 – (1.5)(IQR) = $40,500 – $42,750 = –$2,250
Q3 + (1.5)(IQR) = $69,000 + $42,750 = $111,750
No salary is less than –$2,250. However, $120,000 is more than $11,750, so $120,000 is a potential outlier.
For the two data sets in the test scores example, find the following:
The five number summary for the day and night classes is
| Minimum | Q1 | Median | Q3 | Maximum | |
|---|---|---|---|---|---|
| Day | 32 | 56 | 74.5 | 82.5 | 99 |
| Night | 25.5 | 78 | 81 | 89 | 98 |
The IQR for the night group is Q3 – Q1 = 89 – 78 = 11
The interquartile range (the spread or variability) for the day class is larger than the night class IQR. This suggests more variation will be found in the day class’s class test scores.
Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers.
Night class outliers are calculated as:
For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier.
Find the interquartile range for the following two data sets and compare them.
Test Scores for Class A
Class A
Order the data from smallest to largest.
65 66 67 69 69 76 77 77 79 80 81 83 85 89 90 91 94 96 98 99
IQR = 90.5 – 72.5 = 18
Class B
Order the data from smallest to largest.
68 68 70 71 72 73 75 78 79 80 80 90 90 92 92 95 95 97 99 100
IQR = 93.5 – 72.5 = 21
The data for Class B has a larger IQR, so the scores between Q3 and Q1 (middle 50%) for the data for Class B are more spread out and not clustered about the median.
Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were:
| AMOUNT OF SLEEP PER SCHOOL NIGHT (HOURS) | FREQUENCY | RELATIVE FREQUENCY | CUMULATIVE RELATIVE FREQUENCY |
|---|---|---|---|
| 4 | 2 | 0.04 | 0.04 |
| 5 | 5 | 0.10 | 0.14 |
| 6 | 7 | 0.14 | 0.28 |
| 7 | 12 | 0.24 | 0.52 |
| 8 | 14 | 0.28 | 0.80 |
| 9 | 7 | 0.14 | 0.94 |
| 10 | 3 | 0.06 | 1.00 |
Find the 28th percentile. Notice the 0.28 in the "cumulative relative frequency" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28th percentile is between the last six and the first seven. The 28th percentile is 6.5.
Find the median. Look again at the "cumulative relative frequency" column and find 0.52. The median is the 50th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50th percentile is between the 25th, or seven, and 26th, or seven, values. The median is seven.
Find the third quartile. The third quartile is the same as the 75th percentile. You can "eyeball" this answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75th percentile, then, must be an eight. Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q3, is the 38th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)
Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65th percentile.
| Amount of time spent on route (hours) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 2 | 12 | 0.30 | 0.30 |
| 3 | 14 | 0.35 | 0.65 |
| 4 | 10 | 0.25 | 0.90 |
| 5 | 4 | 0.10 | 1.00 |
The 65th percentile is between the last three and the first four.
The 65th percentile is 3.5.
Using Table 2.25:
Using the data from the frequency table, we have:
Refer to the Table 2.26. Find the third quartile. What is another name for the third quartile?
The third quartile is the 75th percentile, which is four. The 65th percentile is between three and four, and the 90th percentile is between four and 5.75. The third quartile is between 65 and 90, so it must be four.
Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:
If you were to do a little research, you would find several formulas for calculating the kth percentile. Here is one of them.
k = the kth percentile. It may or may not be part of the data.
i = the index (ranking or position of a data value)
n = the total number of data
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
k = 20. Index = i = (29 + 1) = 6. The age in the sixth position is 27. The 20th percentile is 27 years.
k = 55. Index = i = (29 + 1) = 16.5. Round down to 16 and up to 17. The age in the 16th position is 52 and the age in the 17th position is 55. The average of 52 and 55 is 53.5. The 55th percentile is 53.5 years.
You can calculate percentiles using calculators and computers. There are a variety of online calculators.
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
x = 18 and y = 1.(100) = (100) = 63.80. 58 is the 64th percentile.
x = 3 and y = 1.(100) = (100) = 12.07. Twenty-five is the 12th percentile.
Listed are 30 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Percentile for 47: Counting from the bottom of the list, there are 15 data values less than 47. There is one value of 47.
x = 15 and y = 1.(100) = (100) = 53.45. 47 is the 53rd percentile.
Percentile for 31: Counting from the bottom of the list, there are eight data values less than 31. There are two values of 31.
x = 15 and y = 2.(100) = (100) = 31.03. 31 is the 31st percentile.
A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15th percentile.
A percentile may or may not correspond to a value judgment about whether it is "good" or "bad." The interpretation of whether a certain percentile is "good" or "bad" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered "good;" in other contexts a high percentile might be considered "good". In many situations, there is no value judgment that applies.
Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.
When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.
On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.
For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.
Twenty-five percent of runners finished the race in 11.5 seconds or more. Seventy-five percent of runners finished the race in 11.5 seconds or less. A lower percentile is good because finishing a race more quickly is desirable.
On a 20 question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation.
On a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.
Eighty percent of students earned 49 points or fewer. Twenty percent of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable.
At a community college, it was found that the 30th percentile of credit units that students are enrolled for is seven units. Interpret the 30th percentile in the context of this situation.
During a season, the 40th percentile for points scored per player in a game is eight. Interpret the 40th percentile in the context of this situation.
Forty percent of players scored eight points or fewer. Sixty percent of players scored eight points or more. A higher percentile is good because getting more points in a basketball game is desirable.
Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.
0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes
10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;
30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes
Determine the following five values.
If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.
However, the principal needs to be careful. The value 300 appears to be a potential outlier.
Q3 + 1.5(IQR) = 60 + (1.5)(40) = 120.
The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:
We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results.
Box plots (also called box-and-whisker plots or box-whisker plots) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them.
To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The "whiskers" extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data.
You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.
Consider, again, this dataset.
1 1 2 2 4 6 6.8 7.2 8 8.3 9 10 10 11.5
The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot.
See the calculator instructions on the TI web site or in the appendix.
The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line.
It is important to start a box plot with a scaled number line. Otherwise the box plot may not be useful.
The following data are the heights of 40 students in a statistics class.
59 60 61 62 62 63 63 64 64 64 65 65 65 65 65 65 65 65 65 66 66 67 67 68 68 69 70 70 70 70 70 71 71 72 72 73 74 74 75 77
Construct a box plot with the following properties; the calculator intructions for the minimum and maximum values as well as the quartiles follow the example.
To find the minimum, maximum, and quartiles:
Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down.
Put the data values into the list L1.
Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.
Press ENTER.
Use the down and up arrow keys to scroll.
Smallest value = 59.
Largest value = 77.
Q1: First quartile = 64.5.
Q2: Second quartile or median = 66.
Q3: Third quartile = 70.
To construct the box plot:
Press 4:Plotsoff. Press ENTER.
Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER.
Arrow down to Xlist: Press 2nd 1 for L1
Arrow down to Freq: Press ALPHA. Press 1.
Press Zoom. Press 9: ZoomStat.
Press TRACE, and use the arrow keys to examine the box plot.
The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator, and state the interquartile range.
136 140 178 190 205 215 217 218 232 234 240 255 270 275 290 301 303 315 317 318 326 333 343 349 360 369 377 388 391 392 398 400 402 405 408 422 429 450 475 512
IQR = 158
For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like:
In this case, at least 25% of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive.
Test scores for a college statistics class held during the day are:
99 56 78 55.5 32 90 80 81 56 59 45 77 84.5 84 70 72 68 32 79 90
Test scores for a college statistics class held during the evening are:
98 78 68 83 81 89 88 76 65 45 98 90 80 84.5 85 79 78 98 90 79 81 25.5
The following data set shows the heights in inches for the boys in a class of 40 students.
66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74
IQR for the boys = 4
IQR for the girls = 5
The box plot for the heights of the girls has the wider spread for the middle 50% of the data.
Graph a box-and-whisker plot for the data values shown.
1010101535759095100175420490515515790
The five numbers used to create a box-and-whisker plot are:
The following graph shows the box-and-whisker plot.
Follow the steps you used to graph a box-and-whisker plot for the data values shown.
0551530304550506075110140240330
The data are in order from least to greatest. There are 15 values, so the eighth number in order is the median: 50. There are seven data values written to the left of the median and 7 values to the right. The five values that are used to create the boxplot are:
The "center" of a data set is also a way of describing location. The two most widely used measures of the "center" of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.
The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average" is commonly accepted for “arithmetic mean.”
When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an x with a bar over it (pronounced “x bar”): .
The Greek letter μ (pronounced "mew") represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.
To see that both ways of calculating the mean are the same, consider the sample:
In the second example, the frequencies are 3(1) + 2(2) + 1(3) + 5(4).
You can quickly find the location of the median by using the expression .
The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is 97, then = = 49. The median is the 49th value in the ordered data. If the total number of data values is 100, then = = 50.5. The median occurs midway between the 50th and 51st values. The location of the median and the value of the median are not the same. The upper case letter M is often used to represent the median. The next example illustrates the location of the median and the value of the median.
AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):
The calculation for the mean is:
To find the mean and the median:
Clear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER.
Enter data into the list editor. Press STAT 1:EDIT.
Put the data values into list L1.
Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.
Press the down and up arrow keys to scroll.
= 23.6, M = 24
The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median.
3 4 5 7 7 7 7 8 8 9 9 10 10 10 10 10 11 12 12 13 14 14 15 15 17 17 18 19 19 19 21 21 22 22 23 24 24 24 24
Mean: 3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544
Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the "center": the mean or the median?
M = 30,000
(There are 49 people who earn $30,000 and one person who earns $5,000,000.)
The median is a better measure of the "center" than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data.
In a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth $280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median?
The median is the better measure of the “center” than the mean because 59 of the values are $280,000 and one is $2,500,000. The $2,500,000 is an outlier. Either $280,000 or $315,000 gives us a better sense of the middle of the data.
Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.
Statistics exam scores for 20 students are as follows:
5053595963637272727272767881838484849093
Find the mode.
The most frequent score is 72, which occurs five times. Mode = 72.
The number of books checked out from the library from 25 students are as follows:
0001233445577778889101011111212
The most frequent number of books is 7, which occurs four times. Mode = 7.
Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice.
When is the mode the best measure of the "center"? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing.
The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red.
Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software.
Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is $50,000 and the mean is $47,500. What would be the best measure of the “center”?
Because $25,000 occurs nearly half the time, the mode would be the best measure of the center because the median and mean don’t represent what most people make at the factory.
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean of the sample is very likely to get closer and closer to µ. This is discussed in more detail later in the text.
You can think of a sampling distribution as a relative frequency distribution with a great many samples. (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.
| # of movies | Relative Frequency |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 |
If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.
A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean is an example of a statistic which estimates the population mean μ.
When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean: mean = We simply need to modify the definition to fit within the restrictions of a frequency table.
Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is . We can now modify the mean definition to be where f = the frequency of the interval and m = the midpoint of the interval.
A frequency table displaying professor Blount’s last statistic test is shown. Find the best estimate of the class mean.
| Grade Interval | Number of Students |
|---|---|
| 50–56.5 | 1 |
| 56.5–62.5 | 0 |
| 62.5–68.5 | 4 |
| 68.5–74.5 | 4 |
| 74.5–80.5 | 2 |
| 80.5–86.5 | 3 |
| 86.5–92.5 | 4 |
| 92.5–98.5 | 1 |
| Grade Interval | Midpoint |
|---|---|
| 50–56.5 | 53.25 |
| 56.5–62.5 | 59.5 |
| 62.5–68.5 | 65.5 |
| 68.5–74.5 | 71.5 |
| 74.5–80.5 | 77.5 |
| 80.5–86.5 | 83.5 |
| 86.5–92.5 | 89.5 |
| 92.5–98.5 | 95.5 |
Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:
| Hours Teenagers Spend on Video Games | Number of Teenagers |
|---|---|
| 0–3.5 | 3 |
| 3.5–7.5 | 7 |
| 7.5–11.5 | 12 |
| 11.5–15.5 | 7 |
| 15.5–19.5 | 9 |
What is the best estimate for the mean number of hours spent playing video games?
Find the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagers
Consider the following data set.
This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval.
The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median.
The histogram for the data: 4566677778 is not symmetrical. The right-hand side seems "chopped off" compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left.
The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so.
The histogram for the data: 67777888910, is also not symmetrical. It is skewed to the right.
The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most.
To summarize, generally if the distribution of data is skewed to the left, the mean is less than the median, which is often less than the mode. If the distribution of data is skewed to the right, the mode is often less than the median, which is less than the mean.
Skewness and symmetry become important when we discuss probability distributions in later chapters.
Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors.
Terry: 7; 9; 3; 3; 3; 4; 1; 3; 2; 2
Davis: 3; 3; 3; 4; 1; 4; 3; 2; 3; 1
Maris: 2; 3; 4; 4; 4; 6; 6; 6; 8; 3
Discuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center?
a.
b.
| The Ages Former U.S Presidents Died | |
|---|---|
| 4 | 6 9 |
| 5 | 3 6 7 7 7 8 |
| 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |
| 7 | 0 1 1 2 3 4 7 8 8 9 |
| 8 | 0 1 3 5 8 |
| 9 | 0 0 3 3 |
| Key: 8|0 means 80. | |
c.
An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.
Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is five minutes. At supermarket A, the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes.
Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.
Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
Rosa waits for seven minutes:
Binh waits for one minute.
The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because 5 + (1)(2) = 7.
If one were also part of the data set, then one is two standard deviations to the left of five because 5 + (–2)(2) = 1.
The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population.
If x is a number, then the difference "x – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is x – μ. For sample data, in symbols a deviation is x – .
The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ.
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – values for a sample, or the x – μ values for a population). The symbol σ2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample.
In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three.
The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of the Data. How much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is where σ is the standard deviation of the population and n is the size of the sample.
In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation σx or sx from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)
In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of n = 20 fifth grade students. The ages are rounded to the nearest half year:
9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5;
The average age is 10.53 years, rounded to two places.
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s.
| Data | Freq. | Deviations | Deviations2 | (Freq.)(Deviations2) |
|---|---|---|---|---|
| x | f | (x – ) | (x – )2 | (f)(x – )2 |
| 9 | 1 | 9 – 10.525 = –1.525 | (–1.525)2 = 2.325625 | 1 × 2.325625 = 2.325625 |
| 9.5 | 2 | 9.5 – 10.525 = –1.025 | (–1.025)2 = 1.050625 | 2 × 1.050625 = 2.101250 |
| 10 | 4 | 10 – 10.525 = –0.525 | (–0.525)2 = 0.275625 | 4 × 0.275625 = 1.1025 |
| 10.5 | 4 | 10.5 – 10.525 = –0.025 | (–0.025)2 = 0.000625 | 4 × 0.000625 = 0.0025 |
| 11 | 6 | 11 – 10.525 = 0.475 | (0.475)2 = 0.225625 | 6 × 0.225625 = 1.35375 |
| 11.5 | 3 | 11.5 – 10.525 = 0.975 | (0.975)2 = 0.950625 | 3 × 0.950625 = 2.851875 |
| The total is 9.7375 |
The sample variance, s2, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):
The sample standard deviation s is equal to the square root of the sample variance:
which is rounded to two decimal places, s = 0.72.
Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy.
On a baseball team, the ages of each of the players are as follows:
21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40
μ = 30.68
s = 6.09
The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For Example 2.31, there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.
The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.
Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n – 1). Why not divide by n? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n – 1) gives a better estimate of the population variance.
Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.
The standard deviation, s or σ, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large.
The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot.
Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73 – 33 = 40) than the spread in the upper 50% (100 – 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.
| Data | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 33 | 1 | 0.032 | 0.032 |
| 42 | 1 | 0.032 | 0.064 |
| 49 | 2 | 0.065 | 0.129 |
| 53 | 1 | 0.032 | 0.161 |
| 55 | 2 | 0.065 | 0.226 |
| 61 | 1 | 0.032 | 0.258 |
| 63 | 1 | 0.032 | 0.29 |
| 67 | 1 | 0.032 | 0.322 |
| 68 | 2 | 0.065 | 0.387 |
| 69 | 2 | 0.065 | 0.452 |
| 72 | 1 | 0.032 | 0.484 |
| 73 | 1 | 0.032 | 0.516 |
| 74 | 1 | 0.032 | 0.548 |
| 78 | 1 | 0.032 | 0.580 |
| 80 | 1 | 0.032 | 0.612 |
| 83 | 1 | 0.032 | 0.644 |
| 88 | 3 | 0.097 | 0.741 |
| 90 | 1 | 0.032 | 0.773 |
| 92 | 1 | 0.032 | 0.805 |
| 94 | 4 | 0.129 | 0.934 |
| 96 | 1 | 0.032 | 0.966 |
| 100 | 1 | 0.032 | 0.998 (Why isn't this value 1?) |
The following data show the different types of pet food stores in the area carry.
μ = 9.3
s = 2.2
Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:
Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.
Find the standard deviation for the data in Table 2.34.
| Class | Frequency, f | Midpoint, m | m2 | 2 | fm2 | Standard Deviation |
|---|---|---|---|---|---|---|
| 0–2 | 1 | 1 | 1 | 7.58 | 1 | 3.5 |
| 3–5 | 6 | 4 | 16 | 7.58 | 96 | 3.5 |
| 6–8 | 10 | 7 | 49 | 7.58 | 490 | 3.5 |
| 9–11 | 7 | 10 | 100 | 7.58 | 700 | 3.5 |
| 12–14 | 0 | 13 | 169 | 7.58 | 0 | 3.5 |
| 15–17 | 2 | 16 | 256 | 7.58 | 512 | 3.5 |
For this data set, we have the mean, = 7.58 and the standard deviation, sx = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, where sx = sample standard deviation, = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.
Find the standard deviation for the data from the previous example
| Class | Frequency, f |
|---|---|
| 0–2 | 1 |
| 3–5 | 6 |
| 6–8 | 10 |
| 9–11 | 7 |
| 12–14 | 0 |
| 15–17 | 2 |
First, press the STAT key and select 1:Edit
Input the midpoint values into L1 and the frequencies into L2
Select STAT, CALC, and 1: 1-Var Stats
Select 2nd then 1 then , 2nd then 2 Enter
You will see displayed both a population standard deviation, σx, and the sample standard deviation, sx.
The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.
#ofSTDEVs is often called a "z-score"; we can use the symbol z. In symbols, the formulas become:
| Sample | = + zs | |
| Population | = + zσ |
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?
| Student | GPA | School Mean GPA | School Standard Deviation |
|---|---|---|---|
| John | 2.85 | 3.0 | 0.7 |
| Ali | 77 | 80 | 10 |
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.
For John,
For Ali,
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.
John's z-score of –0.21 is higher than Ali's z-score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?
| Swimmer | Time (seconds) | Team Mean Time | Team Standard Deviation |
|---|---|---|---|
| Angie | 26.2 | 27.2 | 0.8 |
| Beth | 27.3 | 30.1 | 1.4 |
For Angie: z = = –1.25
For Beth: z = = –2
The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.
Class Time:
Names:
Collect the Data Record the number of pairs of shoes you own.
| _____ | _____ | _____ | _____ | _____ |
| _____ | _____ | _____ | _____ | _____ |
| _____ | _____ | _____ | _____ | _____ |
| _____ | _____ | _____ | _____ | _____ |
| _____ | _____ | _____ | _____ | _____ |
| _____ | _____ | _____ | _____ | _____ |
A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used.
A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time.
The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50th percentile would be greater than 50 percent of the other obeservations in the set. Quartiles divide data into quarters. The first quartile (Q1) is the 25th percentile,the second quartile (Q2 or median) is 50th percentile, and the third quartile (Q3) is the the 75th percentile. The interquartile range, or IQR, is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q1 from Q3, and can help determine outliers by using the following two expressions.
Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.
The mean and the median can be calculated to help you find the "center" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occuring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set.
Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A right (or positive) skewed distribution has a shape like Figure 2.25. A left (or negative) skewed distribution has a shape like Figure 2.26. A symmetrical distrubtion looks like Figure 2.24.
The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population.
where i = the ranking or position of a data value,
k = the kth percentile,
n = total number of data.
Expression for finding the percentile of a data value: (100)
where x = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,
y = the number of data values equal to the data value for which you want to find the percentile,
n = total number of data
Where f = interval frequencies and m = interval midpoints.
where
For each of the following data sets, create a stem plot and identify any outliers.
The miles per gallon rating for 30 cars are shown below (lowest to highest).
The height in feet of 25 trees is shown below (lowest to highest).
The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.
The data are daily high temperatures in a town for one month.
In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in Table 2.40.
| Number of times in store | Frequency |
|---|---|
| 1 | 4 |
| 2 | 10 |
| 3 | 16 |
| 4 | 6 |
| 5 | 4 |
In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table 2.41.
| Years since last purchase | Frequency |
|---|---|
| 0 | 2 |
| 1 | 8 |
| 2 | 13 |
| 3 | 22 |
| 4 | 16 |
| 5 | 9 |
Several children were asked how many TV shows they watch each day. The results of the survey are shown in Table 2.42.
| Number of TV Shows | Frequency |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 36 |
| 3 | 7 |
| 4 | 2 |
The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table 2.43 shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
| Seasons | Number of students | Proportion of population |
|---|---|---|
| Spring | 8 | 24% |
| Summer | 9 | 26% |
| Autumn | 11 | 32% |
| Winter | 6 | 18% |
Using the data from Mrs. Ramirez’s math class supplied in Exercise 2.8, construct a bar graph showing the percentages.
David County has six high schools. Each school sent students to participate in a county-wide science competition. Table 2.44 shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
| High School | Science competition population | Overall student population |
|---|---|---|
| Alabaster | 28.9% | 8.6% |
| Concordia | 7.6% | 23.2% |
| Genoa | 12.1% | 15.0% |
| Mocksville | 18.5% | 14.3% |
| Tynneson | 24.2% | 10.1% |
| West End | 8.7% | 28.8% |
Use the data from the David County science competition supplied in Exercise 2.10. Construct a bar graph that shows the county-wide population percentage of students at each school.
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
| Data Value (# cars) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
What does the frequency column in Table 2.45 sum to? Why?
What does the relative frequency column in Table 2.45 sum to? Why?
What is the difference between relative frequency and frequency for each data value in Table 2.45?
What is the difference between cumulative relative frequency and relative frequency for each data value?
To construct the histogram for the data in Table 2.45, determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Construct a frequency polygon for the following:
| Pulse Rates for Women | Frequency |
|---|---|
| 60–69 | 12 |
| 70–79 | 14 |
| 80–89 | 11 |
| 90–99 | 1 |
| 100–109 | 1 |
| 110–119 | 0 |
| 120–129 | 1 |
| Actual Speed in a 30 MPH Zone | Frequency |
|---|---|
| 42–45 | 25 |
| 46–49 | 14 |
| 50–53 | 7 |
| 54–57 | 3 |
| 58–61 | 1 |
| Tar (mg) in Nonfiltered Cigarettes | Frequency |
|---|---|
| 10–13 | 1 |
| 14–17 | 0 |
| 18–21 | 15 |
| 22–25 | 7 |
| 26–29 | 2 |
Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.
| Depth of Hunger | Frequency |
|---|---|
| 230–259 | 21 |
| 260–289 | 13 |
| 290–319 | 5 |
| 320–349 | 7 |
| 350–379 | 1 |
| 380–409 | 1 |
| 410–439 | 1 |
Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
| Life Expectancy at Birth – Women | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 3 |
| 77–83 | 8 |
| 84–90 | 2 |
| Life Expectancy at Birth – Men | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 1 |
| 77–83 | 7 |
| 84–90 | 5 |
Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.
| Sex/Year | 1855 | 1856 | 1857 | 1858 | 1859 | 1860 | 1861 |
| Female | 45,545 | 49,582 | 50,257 | 50,324 | 51,915 | 51,220 | 52,403 |
| Male | 47,804 | 52,239 | 53,158 | 53,694 | 54,628 | 54,409 | 54,606 |
| Total | 93,349 | 101,821 | 103,415 | 104,018 | 106,543 | 105,629 | 107,009 |
| Sex/Year | 1862 | 1863 | 1864 | 1865 | 1866 | 1867 | 1868 | 1869 |
| Female | 51,812 | 53,115 | 54,959 | 54,850 | 55,307 | 55,527 | 56,292 | 55,033 |
| Male | 55,257 | 56,226 | 57,374 | 58,220 | 58,360 | 58,517 | 59,222 | 58,321 |
| Total | 107,069 | 109,341 | 112,333 | 113,070 | 113,667 | 114,044 | 115,514 | 113,354 |
| Sex/Year | 1871 | 1870 | 1872 | 1871 | 1872 | 1827 | 1874 | 1875 |
| Female | 56,099 | 56,431 | 57,472 | 56,099 | 57,472 | 58,233 | 60,109 | 60,146 |
| Male | 60,029 | 58,959 | 61,293 | 60,029 | 61,293 | 61,467 | 63,602 | 63,432 |
| Total | 116,128 | 115,390 | 118,765 | 116,128 | 118,765 | 119,700 | 123,711 | 123,578 |
The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973.
| Year | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 |
| Police | 260.35 | 269.8 | 272.04 | 272.96 | 272.51 | 261.34 | 268.89 |
| Homicides | 8.6 | 8.9 | 8.52 | 8.89 | 13.07 | 14.57 | 21.36 |
| Year | 1968 | 1969 | 1970 | 1971 | 1972 | 1973 |
| Police | 295.99 | 319.87 | 341.43 | 356.59 | 376.69 | 390.19 |
| Homicides | 28.03 | 31.49 | 37.39 | 46.26 | 47.24 | 52.33 |
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Listed are 32 ages for Academy Award winning best actors in order from smallest to largest.
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Jesse was ranked 37th in his graduating class of 180 students. At what percentile is Jesse’s ranking?
On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain.
Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th percentile in the context of this situation.
In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain.
In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1,700 in damage and was in the 90th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90th percentile in the context of this problem.
The University of California has two criteria used to set admission standards for freshman to be admitted to a college in the UC system:
Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34th percentile. The 34th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses?
Use Exercise 2.25 to calculate the following values:
First quartile = _______
Second quartile = median = 50th percentile = _______
Third quartile = _______
Interquartile range (IQR) = _____ – _____ = _____
10th percentile = _______
70th percentile = _______
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars.
Construct a box plot below. Use a ruler to measure and scale accurately.
Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas, but not in others? How can you tell?
Find the mean for the following frequency tables.
| Grade | Frequency |
|---|---|
| 49.5–59.5 | 2 |
| 59.5–69.5 | 3 |
| 69.5–79.5 | 8 |
| 79.5–89.5 | 12 |
| 89.5–99.5 | 5 |
| Daily Low Temperature | Frequency |
|---|---|
| 49.5–59.5 | 53 |
| 59.5–69.5 | 32 |
| 69.5–79.5 | 15 |
| 79.5–89.5 | 1 |
| 89.5–99.5 | 0 |
| Points per Game | Frequency |
|---|---|
| 49.5–59.5 | 14 |
| 59.5–69.5 | 32 |
| 69.5–79.5 | 15 |
| 79.5–89.5 | 23 |
| 89.5–99.5 | 2 |
Use the following information to answer the next three exercises: The following data show the lengths of boats moored in a marina. The data are ordered from smallest to largest: 161719202021232425252526262727272829303233333435373940
Calculate the mean.
Identify the median.
Identify the mode.
sample mean = = _______
median = _______
mode = _______
Use the following information to answer the next three exercises: State whether the data are symmetrical, skewed to the left, or skewed to the right.
11122223333333344455
161719222222222223
87878787878889899091
When the data are skewed left, what is the typical relationship between the mean and median?
When the data are symmetrical, what is the typical relationship between the mean and median?
What word describes a distribution that has two modes?
Describe the shape of this distribution.
Describe the relationship between the mode and the median of this distribution.
Describe the relationship between the mean and the median of this distribution.
Describe the shape of this distribution.
Describe the relationship between the mode and the median of this distribution.
Are the mean and the median the exact same in this distribution? Why or why not?
Describe the shape of this distribution.
Describe the relationship between the mode and the median of this distribution.
Describe the relationship between the mean and the median of this distribution.
The mean and median for the data are the same.
345566667777777
Is the data perfectly symmetrical? Why or why not?
Which is the greatest, the mean, the mode, or the median of the data set?
111112121212131517222222
Which is the least, the mean, the mode, and the median of the data set?
5656565859606264646567
Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why?
In a perfectly symmetrical distribution, when would the mode be different from the mean and median?
Use the following information to answer the next two exercises: The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles.
Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth.
Find the value that is one standard deviation below the mean.
Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team?
| Baseball Player | Batting Average | Team Batting Average | Team Standard Deviation |
|---|---|---|---|
| Fredo | 0.158 | 0.166 | 0.012 |
| Karl | 0.177 | 0.189 | 0.015 |
Use Table 2.60 to find the value that is three standard deviations:
Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84.
| Grade | Frequency |
|---|---|
| 49.5–59.5 | 2 |
| 59.5–69.5 | 3 |
| 69.5–79.5 | 8 |
| 79.5–89.5 | 12 |
| 89.5–99.5 | 5 |
| Daily Low Temperature | Frequency |
|---|---|
| 49.5–59.5 | 53 |
| 59.5–69.5 | 32 |
| 69.5–79.5 | 15 |
| 79.5–89.5 | 1 |
| 89.5–99.5 | 0 |
| Points per Game | Frequency |
|---|---|
| 49.5–59.5 | 14 |
| 59.5–69.5 | 32 |
| 69.5–79.5 | 15 |
| 79.5–89.5 | 23 |
| 89.5–99.5 | 2 |
Student grades on a chemistry exam were: 77, 78, 76, 81, 86, 51, 79, 82, 84, 99
Table 2.64 contains the 2010 obesity rates in U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows:
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 10 | |
| 1 | 12 | |
| 2 | 16 | |
| 3 | 12 | |
| 4 | 8 | |
| 5 | 6 | |
| 6 | 2 | |
| 8 | 2 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0–1 | 20 | |
| 2–3 | 35 | |
| 4–5 | 12 | |
| 6–7 | 2 | |
| 8–9 | 1 |
Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 51–100 | 5 | |
| 101–150 | 10 | |
| 151–200 | 15 | |
| 201–250 | 15 | |
| 251–300 | 10 | |
| 301–350 | 5 |
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 100–150 | 5 | |
| 201–250 | 5 | |
| 251–300 | 5 | |
| 301–350 | 5 | |
| 351–400 | 10 | |
| 401–450 | 10 | |
| 451–500 | 10 | |
| 501–550 | 10 | |
| 551–600 | 5 | |
| 601–650 | 5 |
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows.
| # of movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 5 | ||
| 1 | 9 | ||
| 2 | 6 | ||
| 3 | 4 | ||
| 4 | 1 |
Use the following information to answer the next two exercises: Suppose one hundred eleven people who shopped in a special t-shirt store were asked the number of t-shirts they own costing more than $19 each.
The percentage of people who own at most three t-shirts costing more than $19 each is approximately:
If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
Following are the 2010 obesity rates by U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the x-axis with the states.
The median age for U.S. blacks currently is 30.9 years; for U.S. whites it is 42.3 years.
Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in Table 2.72. Also, include left endpoint, but not the right endpoint.
| Salary ($) | Relative Frequency |
|---|---|
| < 20,000 | 0.02 |
| 20,000–25,000 | 0.09 |
| 25,000–30,000 | 0.19 |
| 30,000–40,000 | 0.26 |
| 40,000–50,000 | 0.18 |
| 50,000–75,000 | 0.17 |
| 75,000–99,999 | 0.02 |
| 100,000+ | 0.01 |
Given the following box plot:
The following box plot shows the U.S. population for 1990, the latest available year.
In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results.
Given the following box plot, answer the questions.
Given the following box plots, answer the questions.
A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results.
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
| # of movies | Frequency |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 6 |
| 3 | 4 |
| 4 | 1 |
Construct a box plot of the data.
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table.
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
Table 2.75 gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children?
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years.
A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
75% of all years have an FTES:
The population standard deviation = _____
What percent of the FTES were from 528.5 to 1447.5? How do you know?
What is the IQR? What does the IQR represent?
How many standard deviations away from the mean is the median?
Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here.
| Year | 2005–06 | 2006–07 | 2007–08 | 2008–09 | 2009–10 | 2010–11 |
| Total FTES | 1,585 | 1,690 | 1,735 | 1,935 | 2,021 | 1,890 |
Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR. Round to one decimal place.
Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005.
Compare the IQR for the FTES for 1976–77 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQRs are so different?
Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.
| Student | GPA | School Average GPA | School Standard Deviation |
|---|---|---|---|
| Thuy | 2.7 | 3.2 | 0.8 |
| Vichet | 87 | 75 | 20 |
| Kamala | 8.6 | 8 | 0.4 |
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer.
An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in Table 14.
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
What is the best estimate of the average obesity percentage for these countries? What is the standard deviation for the listed obesity rates? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ obesity rate compared to the average rate? Explain.
Table 2.79 gives the percent of children under five considered to be underweight.
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain.
Santa Clara County, CA, has approximately 27,873 Japanese-Americans. Their ages are as follows:
| Age Group | Percent of Community |
|---|---|
| 0–17 | 18.9 |
| 18–24 | 8.0 |
| 25–34 | 22.8 |
| 35–44 | 15.0 |
| 45–54 | 13.1 |
| 55–64 | 11.9 |
| 65+ | 10.3 |
Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information.
| Javier | Ercilia | |
|---|---|---|
| 6.0 miles | 6.0 miles | |
| 4.0 miles | 7.0 miles |
Use the following information to answer the next three exercises: We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section.
| Number of years | Frequency | Number of years | Frequency |
|---|---|---|---|
| Total = 20 | |||
| 7 | 1 | 22 | 1 |
| 14 | 3 | 23 | 1 |
| 15 | 1 | 26 | 1 |
| 18 | 1 | 40 | 2 |
| 19 | 4 | 42 | 2 |
| 20 | 3 |
What is the IQR?
What is the mode?
Is this a sample or the entire population?
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
| # of movies | Frequency |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 6 |
| 3 | 4 |
| 4 | 1 |
Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows:
| X | Frequency |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 12 |
| 5 | 12 |
| 6 | 0 |
| 7 | 1 |
Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year.
177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265
One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows:
3 8–12 05–31–16 5–2
Refer to Figure 2.58 determine which of the following are true and which are false. Explain your solution to each part in complete sentences.
In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference.
A survey of enrollment at 35 community colleges across the United States yielded the following figures:
6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622
| x | Frequency |
|---|---|
| 0 | 3 |
| 1 | 12 |
| 2 | 33 |
| 3 | 28 |
| 4 | 11 |
| 5 | 9 |
| 6 | 4 |
The 80th percentile is _____
The number that is 1.5 standard deviations BELOW the mean is approximately _____
Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the Table 2.86.
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
Burbary, Ken. Facebook Demographics Revisited – 2001 Statistics, 2011. Available online at http://www.kenburbary.com/2011/03/facebook-demographics-revisited-2011-statistics-2/ (accessed August 21, 2013).
“9th Annual AP Report to the Nation.” CollegeBoard, 2013. Available online at http://apreport.collegeboard.org/goals-and-findings/promoting-equity (accessed September 13, 2013).
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013).
Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker
“Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013).
“Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013).
“Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013).
“Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013).
“CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013).
“Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013).
“Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013).
Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach. CRC Press: 1980.
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013).
Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at http://usatoday30.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1 (accessed April 3, 2013).
Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013).
“1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013).
Data from San Jose Mercury News.
Data from Time Magazine; survey by Yankelovich Partners, Inc.
Data from West Magazine.
Data from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013).
“Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013).
Data from Microsoft Bookshelf.
King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013).
| Stem | Leaf |
|---|---|
| 1 | 9 9 9 |
| 2 | 0 1 1 5 5 5 6 6 8 9 |
| 3 | 1 1 2 2 3 4 5 6 7 7 8 8 8 8 |
| 4 | 1 3 3 |
| Stem | Leaf |
|---|---|
| 2 | 5 5 6 7 7 8 |
| 3 | 0 0 1 2 3 3 5 5 5 7 7 9 |
| 4 | 1 6 9 |
| 5 | 6 7 7 |
| 6 | 1 |
65
The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.
Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values.
Jesse graduated 37th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37.
x = 143 and y = 1. (100) = (100) = 79.72. Jesse’s rank of 37 puts him at the 80th percentile.
When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer.
The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of $1700 or more.
You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more.
4
6 – 4 = 2
6
More than 25% of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25% and the bottom 25% are spread out evenly; the whiskers have the same length.
Mean: 16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 + 34 + 35 + 37 + 39 + 40 = 738;
= 27.33
The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27
4
The data are symmetrical. The median is 3 and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical.
The data are skewed right. The median is 87.5 and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right.
When the data are symmetrical, the mean and median are close or the same.
The distribution is skewed right because it looks pulled out to the right.
The mean is 4.1 and is slightly greater than the median, which is four.
The mode and the median are the same. In this case, they are both five.
The distribution is skewed left because it looks pulled out to the left.
The mean and the median are both six.
The mode is 12, the median is 13.5, and the mean is 15.1. The mean is the largest.
The mean tends to reflect skewing the most because it is affected the most by outliers.
s = 34.5
For Fredo: z = = –0.67
For Karl: z = = –0.8
Fredo’s z-score of –0.67 is higher than Karl’s z-score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.
Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 51–100 | 5 | 0.08 |
| 101–150 | 10 | 0.17 |
| 151–200 | 15 | 0.25 |
| 201–250 | 15 | 0.25 |
| 251–300 | 10 | 0.17 |
| 301–350 | 5 | 0.08 |
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 100–150 | 5 | 0.07 |
| 201–250 | 5 | 0.07 |
| 251–300 | 5 | 0.07 |
| 301–350 | 5 | 0.07 |
| 351–400 | 10 | 0.14 |
| 401–450 | 10 | 0.14 |
| 451–500 | 10 | 0.14 |
| 501–550 | 10 | 0.14 |
| 551–600 | 5 | 0.07 |
| 601–650 | 5 | 0.07 |
c
Answers will vary.
40th percentile will fall between 30,000 and 40,000
80th percentile will fall between 50,000 and 75,000
The mean percentage,
The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or below the median.
474 FTES
919
Hint: Think about the number of years covered by each time period and what happened to higher education during those periods.
For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.
a
b
| Enrollment | Frequency |
|---|---|
| 1000-5000 | 10 |
| 5000-10000 | 16 |
| 10000-15000 | 3 |
| 15000-20000 | 3 |
| 20000-25000 | 1 |
| 25000-30000 | 2 |
a
By the end of this chapter, the student should be able to:
It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.
You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.
Your instructor will survey your class. Count the number of students in the class today.
Use the class data as estimates of the following probabilities. P(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.
Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.
A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes.
An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A).
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition {HT, TH}, so P(A) = = 0.5.
Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P(E) = . If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, of the rolls would result in an outcome of "at least five". You would not expect exactly . The long-term relative frequency of obtaining this result would approach the theoretical probability of as the number of repetitions grows larger and larger.
This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.
"OR" Event:An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.
"AND" Event:An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}.
The complement of event A is denoted A′ (read "A prime"). A′ consists of all outcomes that are NOT in A. Notice that P(A) + P(A′) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A′ = {5, 6}. P(A) = , P(A′) = , and P(A) + P(A′) = = 1
The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is P(A|B) = where P(B) is greater than zero.
For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S).
We get the same result by using the formula. Remember that S has six outcomes.
P(A|B) =
Understanding Terminology and SymbolsIt is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.
The sample space S is the whole numbers starting at one and less than 20.
Let event A = the even numbers and event B = numbers greater than 13.
The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).
A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).
Table 3.1 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed.
| Right-handed | Left-handed | |
|---|---|---|
| Males | 43 | 9 |
| Females | 44 | 4 |
Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities:
Independent and mutually exclusive do not mean the same thing.
Two events are independent if the following are true:
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.
Sampling may be done with replacement or without replacement.
If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.
a. Sampling with replacement:
b. Sampling without replacement:
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.
Which of a. or b. did you sample with replacement and which did you sample without replacement?
a. Without replacement; b. With replacement
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.
without replacement: 1. Possible; 2. Impossible, 3. Possible
with replacement: 1. Possible; 2. Possible, 3. Possible
A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.
For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P(A AND B) = and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive.
If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.
Flip two fair coins. (This is an experiment.)
The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.
Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.
The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is {BB, BR, RB, RR}.
Event A = Getting at least one black card = {BB, BR, RB}
P(A) = = 0.75
Flip two fair coins. Find the probabilities of the events.
Look at the sample space in Example 3.5.
J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive.
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.
Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?
No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = . To be mutually exclusive, P(C AND E) must be zero.
Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P(A) = 0.4 and P(B) = 0.2. P(A AND B) = 0.08. Are events A and B independent? Hint: You must show ONE of the following:
P(A|B) =
The events are independent because P(A|B) = P(A).
Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent?
If G and H are independent, then you must show ONE of the following:
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
a. Show that P(G|H) = P(G).
P(G|H) = = = 0.6 = P(G)
b. Show P(G AND H) = P(G)P(H).
P(G)P(H) = (0.6)(0.5) = 0.3 = P(G AND H)
Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that P(H|G) = P(H) to show that G and H are independent events.
In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
S has ten outcomes. What is P(G AND O)?
Event G and O = {G1, G3}
P(G and O) = = 0.2
Let event C = taking an English class. Let event D = taking a speech class.
Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225.
Justify your answers to the following questions numerically.
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(B AND D) = 0.20.
In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.
Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.
The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.
In a basketball arena,
Let A be the event that a fan is rooting for the away team.
P(B|A) = 0.67
P(B) = 0.25
So P(B) does not equal P(B|A) which means that B and A are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because P(B AND A) = 0.20, not 0.
In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F|L) yet, so you cannot use the second condition.
Solution 1Check whether P(F AND L) = P(F)P(L). We are given that P(F AND L) = 0.45, but P(F)P(L) = (0.60)(0.50) = 0.30. The events of being female and having long hair are not independent because P(F AND L) does not equal P(F)P(L).
Solution 2Check whether P(L|F) equals P(L). We are given that P(L|F) = 0.75, but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.
Interpretation of ResultsThe events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.
Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street.
What is the probability of P(I OR F)?
Because P(I AND F) = 0,
P(I OR F) = P(I) + P(F) - P(I AND F) = 0.44 + 0.56 - 0 = 1
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.
When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.
If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A|B).
This rule may also be written as: P(A|B) =
(The probability of A given B equals the probability of A and B divided by the probability of B.)
If A and B are independent, then P(A|B) = P(A). Then P(A AND B) = P(A|B)P(B) becomes P(A AND B) = P(A)P(B).
If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B).
If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) - P(A AND B) becomes P(A OR B) = P(A) + P(B).
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
a. What is the probability that he makes both goals?
a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90: P(B AND A) = P(B|A) P(A) = (0.90)(0.65) = 0.585
Carlos makes the first and second goals with probability 0.585.
b. What is the probability that Carlos makes either the first goal or the second goal?
b. The problem is asking you to find P(A OR B).
P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715
Carlos makes either the first goal or the second goal with probability 0.715.
c. Are A and B independent?
c. No, they are not, because P(B AND A) = 0.585.
P(B)P(A) = (0.65)(0.65) = 0.423
0.423 ≠ 0.585 = P(B AND A)
So, P(B AND A) is not equal to P(B)P(A).
d. Are A and B mutually exclusive?
d. No, they are not because P(A and B) = 0.585.
To be mutually exclusive, P(A AND B) must equal zero.
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?
P(D|C) = 0.85
P(C AND D) = P(D AND C)
A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.
a. What is the probability that the member is a novice swimmer?
b. What is the probability that the member practices four times a week?
c. What is the probability that the member is an advanced swimmer and practices four times a week?
d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.
e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
e. No, these are not independent events.
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?
Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.
Let: M = math class, S = speech class, M|S = math given speech
a. 0.1625, b. 0.6875, c. No, d. No
A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
a. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
b. Given that the woman has breast cancer, what is the probability that she tests negative?
c. What is the probability that the woman has breast cancer AND tests negative?
d. What is the probability that the woman has breast cancer or tests negative?
e. Are having breast cancer and testing negative independent events?
f. Are having breast cancer and testing negative mutually exclusive?
f. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero.
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?
Let A = student is a senior going to college.
Let B = student plays sports.
P(B) =
P(B|A) =
P(A AND B) = P(B|A)P(A)
P(A AND B) = =
Refer to the information in Example 3.17. P = tests positive.
a. 0.98; b. 0.1401; c. 0.857; d. 0.15
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.
Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:
| Speeding violation in the last year | No speeding violation in the last year | Total | |
|---|---|---|---|
| Cell phone user | 25 | 280 | 305 |
| Not a cell phone user | 45 | 405 | 450 |
| Total | 70 | 685 | 755 |
The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.
Calculate the following probabilities using the table.
a. Find P(Person is a car phone user).
b. Find P(person had no violation in the last year).
c. Find P(Person had no violation in the last year AND was a car phone user).
d. Find P(Person is a car phone user OR person had no violation in the last year).
e. Find P(Person is a car phone user GIVEN person had a violation in the last year).
f. Find P(Person had no violation last year GIVEN person was not a car phone user)
f. (The sample space is reduced to the number of persons who were not car phone users.)
Table 3.3 shows the number of athletes who stretch before exercising and how many had injuries within the past year.
| Injury in last year | No injury in last year | Total | |
|---|---|---|---|
| Stretches | 55 | 295 | 350 |
| Does not stretch | 231 | 219 | 450 |
| Total | 286 | 514 | 800 |
Table 3.4 shows a random sample of 100 hikers and the areas of hiking they prefer.
| Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
|---|---|---|---|---|
| Female | 18 | 16 | ___ | 45 |
| Male | ___ | ___ | 14 | 55 |
| Total | ___ | 41 | ___ | ___ |
a. Complete the table.
a.
| Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
|---|---|---|---|---|
| Female | 18 | 16 | 11 | 45 |
| Male | 16 | 25 | 14 | 55 |
| Total | 34 | 41 | 25 | 100 |
b. Are the events "being female" and "preferring the coastline" independent events?
Let F = being female and let C = preferring the coastline.
Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.
b.
P(F AND C) ≠ P(F)P(C), so the events F and C are not independent.
c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.
c.
d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.
Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.
| Gender | Lake Path | Hilly Path | Wooded Path | Total |
|---|---|---|---|---|
| Female | 45 | 38 | 27 | 110 |
| Male | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is and the probability he is not caught is . If he goes out the second door, the probability he gets caught by Alissa is and the probability he is not caught is . The probability that Alissa catches Muddy coming out of the third door is and the probability she does not catch Muddy is . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is .
| Caught or Not | Door One | Door Two | Door Three | Total |
|---|---|---|---|---|
| Caught | ____ | |||
| Not Caught | ____ | |||
| Total | ____ | ____ | ____ | 1 |
Verify the remaining entries.
a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
a.
| Caught or Not | Door One | Door Two | Door Three | Total |
|---|---|---|---|---|
| Caught | ||||
| Not Caught | ||||
| Total | 1 |
b. What is the probability that Alissa does not catch Muddy?
c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?
c.
Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.
| Year | Robbery | Burglary | Rape | Vehicle | Total |
|---|---|---|---|---|---|
| 2008 | 145.7 | 732.1 | 29.7 | 314.7 | |
| 2009 | 133.1 | 717.7 | 29.1 | 259.2 | |
| 2010 | 119.3 | 701 | 27.7 | 239.1 | |
| 2011 | 113.7 | 702.2 | 26.8 | 229.6 | |
| Total |
TOTAL each column and each row. Total data = 4,520.7
a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575
Table 3.10 relates the weights and heights of a group of individuals participating in an observational study.
| Weight/Height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Obese | 18 | 28 | 14 | |
| Normal | 20 | 51 | 28 | |
| Underweight | 12 | 25 | 9 | |
| Totals |
| Weight/Height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Obese | 18 | 28 | 14 | 60 |
| Normal | 20 | 51 | 28 | 99 |
| Underweight | 12 | 25 | 9 | 46 |
| Totals | 50 | 104 | 51 | 205 |
Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.
A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.
The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:
R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3
The other outcomes are similar.
There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.
a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
a. B1R1 B1R2 B1R3 B2R1 B2R2 B2R3 B3R1 B3R2 B3R3 B4R1 B4R2 B4R3 B5R1 B5R2 B5R3 B6R1 B6R2 B6R3 B7R1 B7R2 B7R3 B8R1 B8R2 B8R3
b. Using the tree diagram, calculate P(RR).
c. Using the tree diagram, calculate P(RB OR BR).
d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).
e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).
e. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = =
This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. = .
f. Using the tree diagram, calculate P(BB).
g. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).
g. P(B on 2nd draw|R on 1st draw) =
There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then .
In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).
Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704.
P(FF) =
An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, .
If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.
a. P(RR) = ________
b. Fill in the blanks:
P(RB OR BR) =
c. P(R on 2nd|B on 1st) =
d. Fill in the blanks.
P(R on 1st AND B on 2nd) = P(RB) = (___)(___) =
e. Find P(BB).
f. Find P(B on 2nd|R on 1st).
f. Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = .
If we are using probabilities, we can label the tree in the following general way.
In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.
A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.
a. c, b. d, c. , d.
Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?
+
A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.
Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:
Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.
Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}.
The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows:
Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.
Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time.
If a student is selected at random, find
Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works.
A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.
The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.
We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.
a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.
In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.
a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks.
b. P(novel or non-fiction) = P(Blue OR Red) = P(Blue) + P(Red) - P(Blue AND Red) = 0.6 + 0.4 - 0.2 = 0.8.
Class time:
Names:
Student Learning Outcomes
Do the Experiment Count out 40 mixed-color M&Ms® which is approximately one small bag’s worth. Record the number of each color in Table 3.12. Use the information from this table to complete Table 3.13. Next, put the M&Ms in a cup. The experiment is to pick two M&Ms, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of Table 3.14. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of Table 3.15. After you record the pick, put both M&Ms back. Do this a total of 24 times, also. Use the data from Table 3.15 to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions.
| Color | Quantity |
|---|---|
| Yellow (Y) | |
| Green (G) | |
| Blue (BL) | |
| Brown (B) | |
| Orange (O) | |
| Red (R) |
| With Replacement | Without Replacement | |
|---|---|---|
| P(2 reds) | ||
| P(R1B2 OR B1R2) | ||
| P(R1 AND G2) | ||
| P(G2|R1) | ||
| P(no yellows) | ||
| P(doubles) | ||
| P(no doubles) |
G2 = green on second pick; R1 = red on first pick; B1 = brown on first pick; B2 = brown on second pick; doubles = both picks are the same colour.
| With Replacement | Without Replacement |
|---|---|
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| ( __ , __ ) ( __ , __ ) | ( __ , __ ) ( __ , __ ) |
| With Replacement | Without Replacement | |
|---|---|---|
| P(2 reds) | ||
| P(R1B2 OR B1R2) | ||
| P(R1 AND G2) | ||
| P(G2|R1) | ||
| P(no yellows) | ||
| P(doubles) | ||
| P(no doubles) |
Discussion Questions
In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.
In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.
The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common.
There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables.
A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.
A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.
A and B are events
P(S) = 1 where S is the sample space
0 ≤ P(A) ≤ 1
P(A|B) =
If A and B are independent, P(A AND B) = P(A)P(B), P(A|B) = P(A) and P(B|A) = P(B).
If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0.
The multiplication rule: P(A AND B) = P(A|B)P(B)
The addition rule: P(A OR B) = P(A) + P(B) - P(A AND B)
In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)
Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti.
Find P(H).
Find P(N).
Find P(F).
Find P(C).
Find P(B).
Find P(G).
Find P(P).
Find P(R).
Find P(Y).
Find P(O).
Find P(A).
Find P(E).
Find P(F).
Find P(N).
Find P(O).
Find P(S).
What is the probability of drawing a red card in a standard deck of 52 cards?
What is the probability of drawing a club in a standard deck of 52 cards?
What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?
What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?
If you land on Y, you get the biggest prize. Find P(Y).
If you land on red, you don’t get a prize. What is P(R)?
Write the symbols for the probability that a player is not an outfielder.
Write the symbols for the probability that a player is an outfielder or is a great hitter.
Write the symbols for the probability that a player is an infielder and is not a great hitter.
Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.
Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.
Write the symbols for the probability that of all the outfielders, a player is not a great hitter.
Write the symbols for the probability that of all the great hitters, a player is an outfielder.
Write the symbols for the probability that a player is an infielder or is not a great hitter.
Write the symbols for the probability that a player is an outfielder and is a great hitter.
Write the symbols for the probability that a player is an infielder.
What is the word for the set of all possible outcomes?
What is conditional probability?
A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book
What is the sum of the probabilities of an event and its complement?
What does P(E|M) mean in words?
What does P(E OR M) mean in words?
E and F are mutually exclusive events. P(E) = 0.4; P(F) = 0.5. Find P(E∣F).
J and K are independent events. P(J|K) = 0.3. Find P(J).
U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find:
Q and R are independent events. P(Q) = 0.4 and P(Q AND R) = 0.1. Find P(R).
Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino.
In this problem, let:
Suppose that one Californian is randomly selected.
Find P(C).
Find P(L).
Find P(C|L).
In words, what is C|L?
Find P(L AND C).
In words, what is L AND C?
Are L and C independent events? Show why or why not.
Find P(L OR C).
In words, what is L OR C?
Are L and C mutually exclusive events? Show why or why not.
Use the following information to answer the next four exercises. Table 3.16 shows a random sample of musicians and how they learned to play their instruments.
| Gender | Self-taught | Studied in School | Private Instruction | Total |
|---|---|---|---|---|
| Female | 12 | 38 | 22 | 72 |
| Male | 19 | 24 | 15 | 58 |
| Total | 31 | 62 | 37 | 130 |
Find P(musician is a female).
Find P(musician is a male AND had private instruction).
Find P(musician is a female OR is self taught).
Are the events “being a female musician” and “learning music in school” mutually exclusive events?
The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation.
Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites.
Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.
| Smoking Level | African American | Native Hawaiian | Latino | Japanese Americans | White | TOTALS |
|---|---|---|---|---|---|---|
| 1–10 | ||||||
| 11–20 | ||||||
| 21–30 | ||||||
| 31+ | ||||||
| TOTALS |
Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.
Find the probability that the person was Latino.
In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.
In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.
In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.
Prove that smoking level/day and ethnicity are dependent events.
The graph in Figure 3.15 displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045.
Explain what is wrong with the following statements. Use complete sentences.
Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score.
Find the probability that an Emotional Health Index Score is 82.7.
Find the probability that an Emotional Health Index Score is 81.0.
Find the probability that an Emotional Health Index Score is more than 81?
Find the probability that an Emotional Health Index Score is between 80.5 and 82?
If we know an Emotional Health Index Score is 81.5 or more, what is the probability that it is 82.7?
What is the probability that an Emotional Health Index Score is 80.7 or 82.7?
What is the probability that an Emotional Health Index Score is less than 80.2 given that it is already less than 81.
What occupation has the highest emotional index score?
What occupation has the lowest emotional index score?
What is the range of the data?
Compute the average EHIS.
If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average EHIS?
On February 28, 2013, a Field Poll Survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them.
In this problem, let:
After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late 2011, the Forum Research polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced:
Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range.
Compute the probability of winning the following types of bets:
Compute the probability of winning the following types of bets:
Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
Roll two fair dice. Each die has six faces.
A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin.
An experiment consists of first rolling a die and then tossing a coin.
An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on.
Consider the following scenario:
Y and Z are independent events.
G and H are mutually exclusive events. P(G) = 0.5 P(H) = 0.3
Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3% speak Spanish.
Let: E = speaks English at home; E′ = speaks another language at home; S = speaks Spanish;
Finish each probability statement by matching the correct answer.
| Probability Statements | Answers |
|---|---|
| a. P(E′) = | i. 0.8043 |
| b. P(E) = | ii. 0.623 |
| c. P(S and E′) = | iii. 0.1957 |
| d. P(S|E′) = | iv. 0.1219 |
1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won green card.
Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned.
Let: R = money returned; E = economics classes; O = other classes
The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected.
| Name | Single | Double | Triple | Home Run | Total Hits |
|---|---|---|---|---|---|
| Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |
| Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |
| Ty Cobb | 3,603 | 174 | 295 | 114 | 4,189 |
| Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |
| Total | 8,471 | 1,577 | 583 | 1,720 | 12,351 |
Are "the hit being made by Hank Aaron" and "the hit being a double" independent events?
United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor.
At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.
In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.
A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student
Use the information in the Table 3.20 to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection.
| Up for reelection: | Democratic Party | Republican Party | Other | Total |
|---|---|---|---|---|
| November 2014 | 20 | 13 | 0 | |
| November 2016 | 10 | 24 | 0 | |
| Total |
What is the probability that a randomly selected senator has an “Other” affiliation?
What is the probability that a randomly selected senator is up for reelection in November 2016?
What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?
What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?
Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?
Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?
The events “Republican” and “Up for reelection in 2016” are ________
The events “Other” and “Up for reelection in November 2016” are ________
Table 3.21 gives the number of suicides estimated in the U.S. for a recent year by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population.
| Race and Sex | 1–14 | 15–24 | 25–64 | over 64 | TOTALS |
|---|---|---|---|---|---|
| white, male | 210 | 3,360 | 13,610 | 22,050 | |
| white, female | 80 | 580 | 3,380 | 4,930 | |
| black, male | 10 | 460 | 1,060 | 1,670 | |
| black, female | 0 | 40 | 270 | 330 | |
| all others | |||||
| TOTALS | 310 | 4,650 | 18,780 | 29,760 |
Do not include "all others" for parts f and g.
Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected.
| NAME | Single | Double | Triple | Home Run | TOTAL HITS |
|---|---|---|---|---|---|
| Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |
| Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |
| Ty Cobb | 3,603 | 174 | 295 | 114 | 4,189 |
| Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |
| TOTAL | 8,471 | 1,577 | 583 | 1,720 | 12,351 |
Find P(hit was made by Babe Ruth).
Find P(hit was made by Ty Cobb|The hit was a Home Run).
Table 3.23 identifies a group of children by one of four hair colors, and by type of hair.
| Hair Type | Brown | Blond | Black | Red | Totals |
|---|---|---|---|---|---|
| Wavy | 20 | 15 | 3 | 43 | |
| Straight | 80 | 15 | 12 | ||
| Totals | 20 | 215 |
In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.
| Shirt# | ≤ 210 | 211–250 | 251–290 | > 290 |
|---|---|---|---|---|
| 1–33 | 21 | 5 | 0 | 0 |
| 34–66 | 6 | 18 | 7 | 4 |
| 66–99 | 6 | 12 | 22 | 5 |
For the following, suppose that you randomly select one player from the 49ers or Cowboys.
Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, P(H) = and P(T) = where H is heads and T is tails.
Find P(tossing a Head on the coin AND a Red bead)
Find P(Blue bead).
A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?)
A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table 3.25.
| Shirt# | ≤ 210 | 211–250 | 251–290 | 290≤ |
|---|---|---|---|---|
| 1–33 | 21 | 5 | 0 | 0 |
| 34–66 | 6 | 18 | 7 | 4 |
| 66–99 | 6 | 12 | 22 | 5 |
For the following, suppose that you randomly select one player from the 49ers or Cowboys.
If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P(Shirt# 1–33|≤ 210 pounds)?
The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a man develops cancer in his lifetime and P = man has at least one false positive.
Given events G and H: P(G) = 0.43; P(H) = 0.26; P(H AND G) = 0.14
Given events J and K: P(J) = 0.18; P(K) = 0.37; P(J OR K) = 0.45
Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.
Suppose that you randomly draw two cards, one at a time, with replacement.
Suppose that you randomly draw two cards, one at a time, without replacement.
Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.
Complete the following.
Suppose that 10,000 U.S. licensed drivers are randomly selected.
Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation.
When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin.
Use the following information to answer the next two exercises. The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories:
| Homosexual/Bisexual | IV Drug User* | Heterosexual Contact | Other | Totals | |
|---|---|---|---|---|---|
| Female | 0 | 70 | 136 | 49 | ____ |
| Male | 2,146 | 463 | 60 | 135 | ____ |
| Totals | ____ | ____ | ____ | ____ | ____ |
Suppose a person with AIDS in Santa Clara County is randomly selected.
Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact.
“Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).
Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013).
Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).
DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013).
Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013).
“Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).
“Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013).
Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013).
Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013).
Data from U.S. Census Bureau.
Data from the Wall Street Journal.
Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013).
Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013).
“Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (accessed May 3, 2013).
Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services.
Data from United States Senate. Available online at www.senate.gov (accessed May 2, 2013).
Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013).
“Human Blood Types.” Unite Blood Services, 2011. Available online at http://www.unitedbloodservices.org/learnMore.aspx (accessed May 2, 2013).
Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013).
“United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013).
Data from Clara County Public H.D.
Data from the American Cancer Society.
Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013).
Data from the Federal Highway Administration, part of the United States Department of Transportation.
Data from the United States Census Bureau, part of the United States Department of Commerce.
Data from USA Today.
“Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013).
“Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html (accessed May 2, 2013).
P(N) = = = 0.36
P(C) = = 0.12
P(G) = = = 0.13
P(R) = = = 0.15
P(O) = = = = 0.11
P(E) = = 0.24
P(N) = = 0.12
P(S) = = = 0.06
= = 0.25
= = 0.5
P(O OR H)
P(H|I)
P(N|O)
P(I OR N)
P(I)
The likelihood that an event will occur given that another event has already occurred.
1
the probability of landing on an even number or a multiple of three
P(J) = 0.3
0.376
C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.
L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.
0.6492
No, because P(L AND C) does not equal 0.
P(musician is a male AND had private instruction) = = = 0.12
P(being a female musician AND learning music in school) = = = 0.29
P(being a female musician)P(learning music in school) = = = = 0.26
No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school).
To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is .
To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is .
0
0.3571
0.2142
Physician (83.7)
83.7 − 79.6 = 4.1
P(Occupation < 81.3) = 0.5
The coin toss is independent of the card picked first.
iii i iv ii
P(type O OR Rh-) = P(type O) + P(Rh-) - P(type O AND Rh-)
0.52 = 0.43 + 0.15 - P(type O AND Rh-); solve to find P(type O AND Rh-) = 0.06
6% of people have type O, Rh- blood
P(NOT(type O AND Rh-)) = 1 - P(type O AND Rh-) = 1 - 0.06 = 0.94
94% of people do not have type O, Rh- blood
0
d
| Race and Sex | 1–14 | 15–24 | 25–64 | over 64 | TOTALS |
|---|---|---|---|---|---|
| white, male | 210 | 3,360 | 13,610 | 4,870 | 22,050 |
| white, female | 80 | 580 | 3,380 | 890 | 4,930 |
| black, male | 10 | 460 | 1,060 | 140 | 1,670 |
| black, female | 0 | 40 | 270 | 20 | 330 |
| all others | 100 | ||||
| TOTALS | 310 | 4,650 | 18,780 | 6,020 | 29,760 |
| Race and Sex | 1–14 | 15–24 | 25–64 | over 64 | TOTALS |
|---|---|---|---|---|---|
| white, male | 210 | 3,360 | 13,610 | 4,870 | 22,050 |
| white, female | 80 | 580 | 3,380 | 890 | 4,930 |
| black, male | 10 | 460 | 1,060 | 140 | 1,670 |
| black, female | 0 | 40 | 270 | 20 | 330 |
| all others | 10 | 210 | 460 | 100 | 780 |
| TOTALS | 310 | 4,650 | 18,780 | 6,020 | 29,760 |
b
a
| <20 | 20–64 | >64 | Totals | |
|---|---|---|---|---|
| Female | 0.0244 | 0.3954 | 0.0661 | 0.486 |
| Male | 0.0259 | 0.4186 | 0.0695 | 0.514 |
| Totals | 0.0503 | 0.8140 | 0.1356 | 1 |
| Car, Truck or Van | Walk | Public Transportation | Other | Totals | |
|---|---|---|---|---|---|
| Alone | 0.7318 | ||||
| Not Alone | 0.1332 | ||||
| Totals | 0.8650 | 0.0390 | 0.0530 | 0.0430 | 1 |
The completed contingency table is as follows:
| Homosexual/Bisexual | IV Drug User* | Heterosexual Contact | Other | Totals | |
|---|---|---|---|---|---|
| Female | 0 | 70 | 136 | 49 | 255 |
| Male | 2,146 | 463 | 60 | 135 | 2,804 |
| Totals | 2,146 | 533 | 196 | 184 | 3,059 |
By the end of this chapter, the student should be able to:
A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%?
Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time?
These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment.
Upper case letters such as X or Y denote a random variable. Lower case letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is given as a number.
For example, let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT; THH; HTH; HHT; HTT; THT; TTH; HHH. Then, x = 0, 1, 2, 3. X is in words and x is a number. Notice that for this example, the x values are countable outcomes. Because you can count the possible values that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is a discrete random variable.
Toss a coin ten times and record the number of heads. After all members of the class have completed the experiment (tossed a coin ten times and counted the number of heads), fill in Table 4.1. Let X = the number of heads in ten tosses of the coin.
| x | Frequency of x | Relative Frequency of x |
|---|---|---|
A discrete probability distribution function has two characteristics:
A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.
P(x) = probability that X takes on a value x.
| x | P(x) |
|---|---|
| 0 | P(x = 0) = |
| 1 | P(x = 1) = |
| 2 | P(x = 2) = |
| 3 | P(x = 3) = |
| 4 | P(x = 4) = |
| 5 | P(x = 5) = |
X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)?
| X | P(x) |
|---|---|
| 0 | P(x = 0) = |
| 1 | P(x = 1) = |
| 2 | P(x = 2) = |
| 3 | P(x = 3) = |
| 4 | P(x = 4) = |
| 5 | P(x = 5) = |
Each P(x) is between 0 and 1, inclusive, and the sum of the probabilities is 1, that is:
Suppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected.
a. Let X = the number of days Nancy ____________________.
b. X takes on what values?
c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.0. The table should have two columns labeled x and P(x). What does the P(x) column sum to?
Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on?
X is the number of days Jeremiah attends basketball practice per week. X takes on the values 0, 1, and 2.
The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.
You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. As you learned in Probability Topics, probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers.
The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ. In other words, after conducting many trials of an experiment, you would expect this average value.
To find the expected value or long term average, μ, simply multiply each value of the random variable by its probability and add the products.
A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, μ, of the number of days per week the men's soccer team plays soccer.
To do the problem, first let the random variable X = the number of days the men's soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column x*P(x). In this column, you will multiply each x value by its probability.
| x | P(x) | x*P(x) |
|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 |
| 1 | 0.5 | (1)(0.5) = 0.5 |
| 2 | 0.3 | (2)(0.3) = 0.6 |
Add the last column x*P(x) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1.
The expected value is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. We say μ = 1.1.
Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight. Calculate the standard deviation of the variable as well.
| x | P(x) | x*P(x) | (x – μ)2 ⋅ P(x) |
|---|---|---|---|
| 0 | P(x = 0) = | (0) = 0 | (0 – 2.1)2 ⋅ 0.04 = 0.1764 |
| 1 | P(x = 1) = | (1) = | (1 – 2.1)2 ⋅ 0.22 = 0.2662 |
| 2 | P(x = 2) = | (2) = | (2 – 2.1)2 ⋅ 0.46 = 0.0046 |
| 3 | P(x = 3) = | (3) = | (3 – 2.1)2 ⋅ 0.18 = 0.1458 |
| 4 | P(x = 4) = | (4) = | (4 – 2.1)2 ⋅ 0.08 = 0.2888 |
| 5 | P(x = 5) = | (5) = | (5 – 2.1)2 ⋅ 0.02 = 0.1682 |
Add the values in the third column of the table to find the expected value of X:
Use μ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value x, multiply the square of its deviation by its probability. (Each deviation has the format x – μ).
Add the values in the fourth column of the table:
0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05
The standard deviation of X is the square root of this sum: σ = ≈ 1.0247
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?
| x | P(x) |
|---|---|
| 0 | P(x = 0) = |
| 1 | P(x = 1) = |
| 2 | P(x = 2) = |
| 3 | P(x = 3) = |
| 4 | P(x = 4) = |
| 5 | P(x = 5) = |
The expected value is 2.24
(0) + (1) + (2) + (3) + (4) + (5) = 0 + + + + + = = 2.24
Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game?
To do this problem, set up an expected value table for the amount of money you can profit.
Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of x are 100,000 dollars and −2 dollars.
To win, you must get all five numbers correct, in order. The probability of choosing one correct number is because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is
Therefore, the probability of winning is 0.00001 and the probability of losing is
The expected value table is as follows:
| x | P(x) | x*P(x) | |
|---|---|---|---|
| Loss | –2 | 0.99999 | (–2)(0.99999) = –1.99998 |
| Profit | 100,000 | 0.00001 | (100000)(0.00001) = 1 |
Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over.
You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term?
Let X = the amount of money you profit. The x-values are –$1 and $256.
The probability of guessing the right suit each time is = 0.0039
The probability of losing is 1 – = = 0.9961
(0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023 or 0.23 cents.
Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = and P(tails) = . If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, will you come out ahead?
a. Define a random variable X.
b. Complete the following expected value table.
| x | ____ | ____ | |
|---|---|---|---|
| WIN | 10 | ____ | |
| LOSE | ____ | ____ |
c. What is the expected value, μ? Do you come out ahead?
c. Add the last column of the table. The expected value μ = . You lose, on average, about 67 cents each time you play the game so you do not come out ahead.
Suppose you play a game with a spinner. You play each game by spinning the spinner once. P(red) = , P(blue) = , and P(green) = . If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win $10. Complete the following expected value table.
| x | P(x) | ||
|---|---|---|---|
| Red | |||
| Blue | |||
| Green | 10 |
| x | P(x) | x*P(x) | |
|---|---|---|---|
| Red | –10 | ||
| Blue | 0 | ||
| Green | 10 |
Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled (x – μ)2P(x) and take the square root.
| x | P(x) | x*P(x) | (x – μ)2P(x) |
|---|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 | (0 – 1.1)2(0.2) = 0.242 |
| 1 | 0.5 | (1)(0.5) = 0.5 | (1 – 1.1)2(0.5) = 0.005 |
| 2 | 0.3 | (2)(0.3) = 0.6 | (2 – 1.1)2(0.3) = 0.243 |
Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root of 0.49, or σ = = 0.7
Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating μ and σ.
Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like Table 4.11 and calculate the mean μ and standard deviation σ of X.
Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:
| (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |
| (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |
| (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |
| (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |
| (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |
| (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |
Use the sample space to complete the following table:
| x | P(x) | xP(x) | (x – μ)2 P(x) |
|---|---|---|---|
| 0 | 0 | (0 – 1)2 ⋅ = | |
| 1 | (1 – 1)2 ⋅ = 0 | ||
| 2 | (1 – 1)2 ⋅ = |
Add the values in the third column to find the expected value: μ = = 1. Use this value to complete the fourth column.
Add the values in the fourth column and take the square root of the sum: σ = ≈ 0.7071.
On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay $20. Let X = the amount of profit from a bet.
P(win) = P(one moderate earthquake will occur) = 21.42%
P(loss) = P(one moderate earthquake will not occur) = 100% – 21.42%
If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of X? Construct a table similar to Table 4.13 and Table 4.14 to help you answer these questions.
| x | P(x) | x(Px) | (x – μ)2P(x) | |
|---|---|---|---|---|
| win | 50 | 0.2142 | 10.71 | [50 – (–5.006)]2(0.2142) = 648.0964 |
| loss | –20 | 0.7858 | –15.716 | [–20 – (–5.006)]2(0.7858) = 176.6636 |
Mean = Expected Value = 10.71 + (–15.716) = –5.006.
If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5.01 per bet.
On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. As in Example 4.7, you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let X = the amount of profit from a bet. Find the mean and standard deviation of X.
| x | P(x) | x ⋅ (Px) | (x - μ)2 P(x) | |
|---|---|---|---|---|
| win | 100 | 0.0108 | 1.08 | [100 – (–8.812)]2 ⋅ 0.0108 = 127.8726 |
| loss | –10 | 0.9892 | –9.892 | [–10 – (–8.812)]2 ⋅ 0.9892 = 1.3961 |
Mean = Expected Value = μ = 1.08 + (–9.892) = –8.812
If you make this bet many times under the same conditions, your long term outcome will be an average loss of $8.81 per bet.
Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions.
A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions.
There are three characteristics of a binomial experiment.
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials.
The mean, μ, and variance, σ2, for the binomial probability distribution are μ = np and σ2 = npq. The standard deviation, σ, is then σ = .
Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.
At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. The random variable X = the number of students who withdraw from the randomly selected elementary physics class.
The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?
a school that offers fruit in their lunch every day
Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can be stated mathematically as P(x = 15).
A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.
P(x = 12)
A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let X = the number of heads in 15 flips of the fair coin. X takes on the values 0, 1, 2, 3, ..., 15. Since the coin is fair, p = 0.5 and q = 0.5. The number of trials is n = 15. State the probability question mathematically.
P(x > 10)
A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.
P(x > 3)
Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.
a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.
b. If we are interested in the number of students who do their homework on time, then how do we define X?
c. What values does x take on?
d. What is a "failure," in words?
d. Failure is defined as a student who does not complete his or her homework on time.
The probability of a success is p = 0.70. The number of trials is n = 50.
e. If p + q = 1, then what is q?
f. The words "at least" translate as what kind of inequality for the probability question P(x ____ 40).
Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem.
This is a binomial problem because there is only a success or a failure, and there are a definite number of trials. The probability of a success stays the same for each trial.
X ~ B(n, p)
Read this as "X is a random variable with a binomial distribution." The parameters are n and p; n = number of trials, p = probability of a success on each trial.
It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?
Let X = the number of workers who have a high school diploma but do not pursue any further education.
X takes on the values 0, 1, 2, ..., 20 where n = 20, p = 0.41, and q = 1 – 0.41 = 0.59. X ~ B(20, 0.41)
Find P(x ≤ 12). P(x ≤ 12) = 0.9738. (calculator or computer)
Go into 2nd DISTR. The syntax for the instructions are as follows:
To calculate (x = value): binompdf(n, p, number) if "number" is left out, the result is the binomial probability table.
If you want to find P(x = 12), use the pdf (binompdf). If you want to find P(x > 12), use 1 - binomcdf(20,0.41,12).
The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.
The graph of X ~ B(20, 0.41) is as follows:
The y-axis contains the probability of x, where X = the number of workers who have only a high school diploma.
The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(0.41) = 8.2.
The formula for the variance is σ2 = npq. The standard deviation is σ = .
About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.
P(x ≤ 14) = 0.9695
In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists.
According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending.
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X = the number of people who will develop pancreatic cancer.
During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X = the number of shots that scored points.
The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is . The probability of a student on the second draw is , when the first draw selects a student. The probability is , when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.
A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.
This is not binomial because the names are not replaced, which means the probability changes for each time a name is drawn. This violates the condition of independence.
There are three main characteristics of a geometric experiment.
X = the number of independent trials until the first success.
You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P(x = 5).
You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on?
1, 2, 3, 4, … n. It can go on indefinitely.
A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions?
Let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). ("At least" translates to a "greater than or equal to" symbol).
An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically?
P(x ≥ 10)
Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?
This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).
a. Let X = the number of ____________ you must ask ____________ one says yes.
b. What values does X take on?
c. What are p and q?
d. The probability question is P(_______).
d. P(x = 4)
You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q?
p = 0.1
q = 0.9
X ~ G(p)
Read this as "X is a random variable with a geometric distribution." The parameter is p; p = the probability of a success for each trial.
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Let X = the number of computer components tested until the first defect is found.
X takes on the values 1, 2, 3, ... where p = 0.02. X ~ G(0.02)
Find P(x = 7). P(x = 7) = 0.0177.
To find the probability that x = 7,
To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function.
The probability that the seventh component is the first defect is 0.0177.
The graph of X ~ G(0.02) is:
The y-axis contains the probability of x, where X = the number of computer components tested.
The number of components that you would expect to test until you find the first defective one is the mean, .
The formula for the mean is μ = = = 50
The formula for the variance is σ2 = = = 2,450
The standard deviation is σ = = = 49.5
The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.
P(x = 9) = 0.0092
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G or X ~ G(0.0128).
The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the number of Afghani women you ask until one says that she is literate.
There are five characteristics of a hypergeometric experiment.
The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest.
A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let X = the number of gumdrops in the sample of 50. X takes on the values x = 0, 1, 2, ..., 50. What is the probability statement written mathematically?
P(x = 35)
A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample?
The group of interest is the vowel letter tiles. The size of the group of interest is 44. The size of the sample is seven.
Suppose a shipment of 100 DVD players is known to have ten defective players. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 players, at most two are defective. The two groups are the 90 non-defective DVD players and the 10 defective DVD players. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective DVD players. The size of the sample is 12 DVD players. (They may be non-defective or defective.) Let X = the number of defective DVD players in the sample of 12. X takes on the values 0, 1, 2, ..., 10. X may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective DVD players. Write the probability statement mathematically.
P(x ≤ 2)
A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is X, and what values does it take on?
Let X = the number of cracked eggs in the sample of 15. X takes on the values 0, 1, 2, …, 12.
You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men?
This is a hypergeometric problem because you are choosing your committee from two groups (men and women).
a. Are you choosing with or without replacement?
b. What is the group of interest?
c. How many are in the group of interest?
d. How many are in the other group?
e. Let X = _________ on the committee. What values does X take on?
f. The probability question is P(_______).
f. P(x > 4)
A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem.
X ~ H(r, b, n)
Read this as "X is a random variable with a hypergeometric distribution." The parameters are r, b, and n; r = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen sample.
A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee?
Let X = the number of men on the committee of four. The men are the group of interest (first group).
X takes on the values 0, 1, 2, 3, 4, where r = 6, b = 5, and n = 4. X ~ H(6, 5, 4)
Find P(x = 2). P(x = 2) = 0.4545 (calculator or computer)
Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do.
The probability that there are two men on the committee is about 0.45.
The graph of X ~ H(6, 5, 4) is:
The y-axis contains the probability of X, where X = the number of men on the committee.
You would expect m = 2.18 (about two) men on the committee.
The formula for the mean is
An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has ten slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample?
The group of interest is the 15 boys. The sample consists of the ten slots on the intramural basketball team.
There are two main characteristics of a Poisson experiment.
The random variable X = the number of occurrences in the interval of interest.
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?
Let X = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is (12) = 2 loaves of bread.
The probability question asks you to find P(x = 3).
The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes?
(8) = 2 fish
A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.
P(x < 5)
An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically.
P(x < 8)
You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast.
This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.
a. What is the interval of interest?
b. What is the average number of times the news reporter says "uh" during one broadcast?
c. Let X = ____________. What values does X take on?
c. Let X = the number of times the news reporter says "uh" during one broadcast.
d. The probability question is P(______).
d. P(x > 2)
An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution.
This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate. The events are independent.
X ~ P(μ)
Read this as "X is a random variable with a Poisson distribution." The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest.
Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or hour.)
x = 0, 1, 2, 3, ...
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives
(6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem.
X ~ P(0.75)
Find P(x > 1). P(x > 1) = 0.1734 (calculator or computer)
The TI calculators use λ (lambda) for the mean.
The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:
The graph of X ~ P(0.75) is:
The y-axis contains the probability of x where X = the number of calls in 15 minutes.
A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer.
P(x > 4) = 0.0527
According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(147). The mean is 147 emails.
According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(187). The mean is 187 text messages.
Text message users receive or send an average of 41.5 text messages per day.
Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.
The Poisson distribution can be used to approximate probabilities for a binomial distribution. This next example demonstrates the relationship between the Poisson and the binomial distributions. Let n represent the number of binomial trials and let p represent the probability of a success for each trial. If n is large enough and p is small enough then the Poisson approximates the binomial very well. In general, n is considered “large enough” if it is greater than or equal to 20. The probability p from the binomial distribution should be less than or equal to 0.05. When the Poisson is used to approximate the binomial, we use the binomial mean μ = np. The variance of X is σ2 = μ and the standard deviation is σ = . The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.
On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Let X = the number of days with low seismic activity.
Using the binomial distribution:
Using the Poisson distribution:
We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0.
On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Let X = the number of days with moderate seismic activity.
Using the binomial distribution: P(x = 5) = binompdf(100, 0.0143, 5) ≈ 0.0115
Using the Poisson distribution:
We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—the difference between the values is 0.0004.
Class Time:
Names:
ProcedureThe experimental procedure is to pick one card from a deck of shuffled cards.
| x | Frequency | Relative Frequency |
|---|---|---|
| 0 | __________ | __________ |
| 1 | __________ | __________ |
| 2 | __________ | __________ |
| 3 | __________ | __________ |
| 4 | __________ | __________ |
| 5 | __________ | __________ |
| 6 | __________ | __________ |
| 7 | __________ | __________ |
| 8 | __________ | __________ |
| 9 | __________ | __________ |
| 10 | __________ | __________ |
| x | P(x) |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 | |
| 7 | |
| 8 | |
| 9 | |
| 10 |
Using the Data
RF = relative frequency
Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.
Discussion QuestionsFor questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
Class Time:
Names:
Procedure
Organize the DataIn Table 4.20, fill in the y value that corresponds to each x value. Next, record the number of matches picked for your class. Then, calculate the relative frequency.
| x | y | Frequency | Relative Frequency |
|---|---|---|---|
| 0 | |||
| 1 | |||
| 2 | |||
| 3 |
Theoretical DistributionBuild the theoretical PDF chart for x and y based on the distribution from the Procedure section.
| x | y | P(x) = P(y) |
|---|---|---|
| 0 | ||
| 1 | ||
| 2 | ||
| 3 |
Use the Data
RF = relative frequency
Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.
Discussion QuestionFor questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows:
The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.
A statistical experiment can be classified as a binomial experiment if the following conditions are met:
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np, and the standard deviation is given by the formula σ = .
There are three characteristics of a geometric experiment:
In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G(p) where p is the probability of success in a single trial.
The mean of the geometric distribution X ~ G(p) is μ = = .
A hypergeometric experiment is a statistical experiment with the following properties:
The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. The distribution of X is denoted X ~ H(r, b, n), where r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. It follows that
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is "small" (less than or equal to 0.05) and the number of trials is "large" (greater than or equal to 20).
Mean or Expected Value:
Standard Deviation:
X ~ B(n, p) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p.
X = the number of successes in n independent trials
n = the number of independent trials
X takes on the values x = 0, 1, 2, 3, ..., n
p = the probability of a success for any trial
q = the probability of a failure for any trial
p + q = 1
q = 1 – p
The mean of X is μ = np. The standard deviation of X is σ = .
X ~ G(p) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p.
X = the number of independent trials until the first success
X takes on the values x = 1, 2, 3, ...
p = the probability of a success for any trial
q = the probability of a failure for any trial p + q = 1
The mean is μ = .
The standard deviation is σ = = .
X ~ H(r, b, n) means that the discrete random variable X has a hypergeometric probability distribution with r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample.
X = the number of items from the group of interest that are in the chosen sample, and X may take on the values x = 0, 1, ..., up to the size of the group of interest. (The minimum value for X may be larger than zero in some instances.)
n ≤ r + b
The mean of X is given by the formula μ = and the standard deviation is = .
X ~ P(μ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest.
X takes on the values x = 0, 1, 2, 3, ...
The mean μ is typically given.
The variance is σ2 = μ, and the standard deviation is
When P(μ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial.
Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution.
Let X = the number of years a new hire will stay with the company.
Let P(x) = the probability that a new hire will stay with the company x years.
Complete Table 4.22 using the data provided.
| x | P(x) |
|---|---|
| 0 | 0.12 |
| 1 | 0.18 |
| 2 | 0.30 |
| 3 | 0.15 |
| 4 | |
| 5 | 0.10 |
| 6 | 0.05 |
P(x = 4) = _______
P(x ≥ 5) = _______
On average, how long would you expect a new hire to stay with the company?
What does the column “P(x)” sum to?
| x | P(x) |
|---|---|
| 1 | 0.15 |
| 2 | 0.35 |
| 3 | 0.40 |
| 4 | 0.10 |
Define the random variable X.
What is the probability the baker will sell more than one batch? P(x > 1) = _______
What is the probability the baker will sell exactly one batch? P(x = 1) = _______
On average, how many batches should the baker make?
Define the random variable X.
Construct a probability distribution table for the data.
We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic?
Define the random variable X.
What values does x take on?
Construct a PDF table.
Find the probability that Javier volunteers for less than three events each month. P(x < 3) = _______
Find the probability that Javier volunteers for at least one event each month. P(x > 0) = _______
Complete the expected value table.
| x | P(x) | x*P(x) |
|---|---|---|
| 0 | 0.2 | |
| 1 | 0.2 | |
| 2 | 0.4 | |
| 3 | 0.2 |
Find the expected value from the expected value table.
| x | P(x) | x*P(x) |
|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 |
| 4 | 0.3 | 4(0.3) = 1.2 |
| 6 | 0.4 | 6(0.4) = 2.4 |
| 8 | 0.2 | 8(0.2) = 1.6 |
Find the standard deviation.
| x | P(x) | x*P(x) | (x – μ)2P(x) |
|---|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 | (2–5.4)2(0.1) = 1.156 |
| 4 | 0.3 | 4(0.3) = 1.2 | (4–5.4)2(0.3) = 0.588 |
| 6 | 0.4 | 6(0.4) = 2.4 | (6–5.4)2(0.4) = 0.144 |
| 8 | 0.2 | 8(0.2) = 1.6 | (8–5.4)2(0.2) = 1.352 |
Identify the mistake in the probability distribution table.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.50 |
| 3 | 0.30 | 0.90 |
| 4 | 0.20 | 0.80 |
| 5 | 0.15 | 0.75 |
Identify the mistake in the probability distribution table.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.40 |
| 3 | 0.25 | 0.65 |
| 4 | 0.20 | 0.85 |
| 5 | 0.15 | 1 |
Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.35 | |
| 2 | 0.20 | |
| 3 | 0.15 | |
| 4 | ||
| 5 | 0.10 | |
| 6 | 0.05 |
Define the random variable X.
Define P(x), or the probability of x.
Find the probability that a physics major will do post-graduate research for four years. P(x = 4) = _______
FInd the probability that a physics major will do post-graduate research for at most three years. P(x ≤ 3) = _______
On average, how many years would you expect a physics major to spend doing post-graduate research?
Complete Table 4.30 using the data provided.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.10 | |
| 2 | 0.05 | |
| 3 | 0.10 | |
| 4 | ||
| 5 | 0.30 | |
| 6 | 0.20 | |
| 7 | 0.10 |
In words, define the random variable X.
P(x = 4) = _______
P(x < 4) = _______
On average, how many years would you expect a child to study ballet with this teacher?
What does the column "P(x)" sum to and why?
What does the column "x*P(x)" sum to and why?
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game?
Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status.
In words, define the random variable X.
X ~ _____(_____,_____)
What values does the random variable X take on?
Construct the probability distribution function (PDF).
| x | P(x) |
|---|---|
On average (μ), how many would you expect to answer yes?
What is the standard deviation (σ)?
What is the probability that at most five of the freshmen reply “yes”?
What is the probability that at least two of the freshmen reply “yes”?
Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask.
In words, define the random variable X.
X ~ _____(_____,_____)
What values does the random variable X take on?
Construct the probability distribution function (PDF). Stop at x = 6.
| x | P(x) |
|---|---|
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 |
On average (μ), how many freshmen would you expect to have to ask until you found one who replies "yes?"
What is the probability that you will need to ask fewer than three freshmen?
Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample.
In words, define the random variable X.
X ~ _____(_____,_____)
What values does X take on?
Find the standard deviation.
On average (μ), how many would you expect to be business majors?
Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.
Assume the event occurs independently in any given day. Define the random variable X.
What values does X take on?
What is the probability of getting 150 customers in one day?
What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.
What is the probability that the store will have more than 12 customers in the first hour?
What is the probability that the store will have fewer than 12 customers in the first two hours?
Which type of distribution can the Poisson model be used to approximate? When would you do this?
Assume the event occurs independently in any given day. In words, define the random variable X.
X ~ _____(_____,_____)
What values does X take on?
For the given values of the random variable X, fill in the corresponding probabilities.
Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically.
Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.
Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in Table 4.33.
| x | P(x) |
|---|---|
| 3 | 0.05 |
| 4 | 0.40 |
| 5 | 0.30 |
| 6 | 0.15 |
| 7 | 0.10 |
A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150.
A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.
You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss.
Complete the PDF and answer the questions.
| x | P(x) | xP(x) |
|---|---|---|
| 0 | 0.3 | |
| 1 | 0.2 | |
| 2 | ||
| 3 | 0.4 |
Suppose that you are offered the following “deal.” You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6.
A venture capitalist, willing to invest $1,000,000, has three investments to choose from. The first investment, a software company, has a 10% chance of returning $5,000,000 profit, a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning $3,000,000 profit, a 40% chance of returning $1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning $6,000,000 profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars.
Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have.
| x | P(x) | xP(x) |
|---|---|---|
| 0 | 0.10 | |
| 1 | 0.20 | |
| 2 | 0.30 | |
| 3 | ||
| 4 | 0.10 | |
| 5 | 0.05 | |
| 6 (or more) | 0.05 |
Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in Table 4.36.
| x | P(x) |
|---|---|
| 3 | 0.05 |
| 4 | 0.40 |
| 5 | 0.30 |
| 6 | 0.15 |
| 7 | 0.10 |
On average, how many years do you expect it to take for an individual to earn a B.S.?
People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.50 |
| 2 | 0.24 |
| 3 | |
| 4 | 0.70 |
| 5 | 0.04 |
| x | P(x) |
|---|---|
| 0 | 0.35 |
| 1 | 0.25 |
| 2 | 0.20 |
| 3 | 0.10 |
| 4 | 0.05 |
| 5 | 0.05 |
A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.
Based upon the financial gain or loss over the long run, should you play the game?
Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.
In a lottery, there are 250 prizes of $5, 50 prizes of $25, and ten prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even?
According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery.
Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf.
Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu.
Define the random variable and list its possible values.
State the distribution of X.
Find the probability that at least four of the 25 patients actually have the flu.
On average, for every 25 patients calling in, how many do you expect to have the flu?
People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given Table 4.39. There is five-video limit per customer at this store, so nobody ever rents more than five DVDs.
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.50 |
| 2 | 0.24 |
| 3 | |
| 4 | 0.07 |
| 5 | 0.04 |
A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.
Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games.
The expected number of wins for that upcoming month is:
Let X = the number of games won in that upcoming month.
What is the probability that the San Jose Sharks win six games in that upcoming month?
What is the probability that the San Jose Sharks win at least five games in that upcoming month
A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct.
A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly.
Six different colored dice are rolled. Of interest is the number of dice that show a one.
More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses.
Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen.
At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon.
Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school.
The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent.
It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies.
There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The “house” rolls three dice. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back his or her $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game.
According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = the number of people who have access to electricity.
The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people in Afghanistan at random. Let X = the number of people who are literate.
A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.
Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl.
It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.
In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.
Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.
Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on?
The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are infected with HIV.”1 In South Africa, the prevalence of HIV is 17.3%. Let X = the number of people you test until you find a person infected with HIV.
According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millenials you ask until you find a person without a profile on a social networking site.
A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration.
In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once.
Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient.
Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked.
A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart?
The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon.
The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour.
A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.
The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.
The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.
Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:
The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.
According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.
The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions.
Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.
On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.
Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.
In words, the random variable X = _________________
Find the probability that her cats will wake her up no more than five times next week.
Class Catalogue at the Florida State University. Available online at https://apps.oti.fsu.edu/RegistrarCourseLookup/SearchFormLegacy (accessed May 15, 2013).
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| x | P(x) |
|---|---|
| 0 | 0.12 |
| 1 | 0.18 |
| 2 | 0.30 |
| 3 | 0.15 |
| 4 | 0.10 |
| 5 | 0.10 |
| 6 | 0.05 |
0.10 + 0.05 = 0.15
1
0.35 + 0.40 + 0.10 = 0.85
1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.04 |
| 2 | 0.08 |
| 3 | 0.85 |
Let X = the number of events Javier volunteers for each month.
| x | P(x) |
|---|---|
| 0 | 0.05 |
| 1 | 0.05 |
| 2 | 0.10 |
| 3 | 0.20 |
| 4 | 0.25 |
| 5 | 0.35 |
1 – 0.05 = 0.95
0.2 + 1.2 + 2.4 + 1.6 = 5.4
The values of P(x) do not sum to one.
Let X = the number of years a physics major will spend doing post-graduate research.
1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15
1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years
X is the number of years a student studies ballet with the teacher.
0.10 + 0.05 + 0.10 = 0.25
The sum of the probabilities sum to one because it is a probability distribution.
X = the number that reply “yes”
0, 1, 2, 3, 4, 5, 6, 7, 8
5.7
0.4151
X = the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.
1,2,…
1.4
X = the number of business majors in the sample.
2, 3, 4, 5, 6, 7, 8, 9
6.26
0, 1, 2, 3, 4, …
0.0485
0.0214
X = the number of U.S. teens who die from motor vehicle injuries per day.
0, 1, 2, 3, 4, ...
No
The variable of interest is X, or the gain or loss, in dollars.
The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.
We first need to construct the probability distribution for X. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value.
| Card Event | X net gain/loss | P(X) |
|---|---|---|
| Face Card and Heads | 6 | |
| Face Card and Tails | 2 | |
| (Not Face Card) and (H or T) | –2 |
| Software Company | |
|---|---|
| x | P(x) |
| 5,000,000 | 0.10 |
| 1,000,000 | 0.30 |
| –1,000,000 | 0.60 |
| Hardware Company | |
|---|---|
| x | P(x) |
| 3,000,000 | 0.20 |
| 1,000,000 | 0.40 |
| –1,000,00 | 0.40 |
| Biotech Firm | |
|---|---|
| x | P(x) |
| 6,00,000 | 0.10 |
| 0 | 0.70 |
| –1,000,000 | 0.20 |
4.85 years
b
Let X = the amount of money to be won on a ticket. The following table shows the PDF for X.
| x | P(x) |
|---|---|
| 0 | 0.969 |
| 5 | = 0.025 |
| 25 | = 0.005 |
| 100 | = 0.001 |
Calculate the expected value of X.
0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35
A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.
X = the number of patients calling in claiming to have the flu, who actually have the flu.
X = 0, 1, 2, ...25
0.0165
d. 4.43
c
0, 1, 2, and 3
Let X = the number of defective bulbs in a string.
Using the Poisson distribution:
Using the binomial distribution:
The Poisson approximation is very good—the difference between the probabilities is only 0.0026.
d
By the end of this chapter, the student should be able to:
Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables.
The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important.
The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve.
The curve is called the probability density function (abbreviated as pdf). We use the symbol f(x) to represent the curve. f(x) is the function that corresponds to the graph; we use the density function f(x) to draw the graph of the probability distribution.
Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf). The cumulative distribution function is used to evaluate probability as area.
We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus. However, because most students taking this course have not studied calculus, we will not be using calculus in this textbook.
There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way.
In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions.
We begin by defining a continuous probability density function. We use the function notation f(x). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f(x) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA.
Consider the function f(x) = for 0 ≤ x ≤ 20. x = a real number. The graph of f(x) = is a horizontal line. However, since 0 ≤ x ≤ 20, f(x) is restricted to the portion between x = 0 and x = 20, inclusive.
f(x) = for 0 ≤ x ≤ 20.
The graph of f(x) = is a horizontal line segment when 0 ≤ x ≤ 20.
The area between f(x) = where 0 ≤ x ≤ 20 and the x-axis is the area of a rectangle with base = 20 and height = .
Suppose we want to find the area between f(x) = and the x-axis where 0 < x < 2.
area of a rectangle = (base)(height).
The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as P(0 < x < 2) = P(x < 2) = 0.1.
Suppose we want to find the area between f(x) = and the x-axis where 4 < x < 15.
The area corresponds to the probability P(4 < x < 15) = 0.55.
Suppose we want to find P(x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P(x = 15) = (base)(height) = (0) = 0
P(X ≤ x) (can be written as P(X < x) for continuous distributions) is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can use the CDF to calculate P(X > x). The CDF gives "area to the left" and P(X > x) gives "area to the right." We calculate P(X > x) for continuous distributions as follows: P(X > x) = 1 – P (X < x).
Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = , 0 ≤ x ≤ 20.
To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle.
Consider the function f(x) = for 0 ≤ x ≤ 8. Draw the graph of f(x) and find P(2.5 < x < 7.5).
P (2.5 < x < 7.5) = 0.625
The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.
The data in Table 5.1 are 55 smiling times, in seconds, of an eight-week-old baby.
| 10.4 | 19.6 | 18.8 | 13.9 | 17.8 | 16.8 | 21.6 | 17.9 | 12.5 | 11.1 | 4.9 |
| 12.8 | 14.8 | 22.8 | 20.0 | 15.9 | 16.3 | 13.4 | 17.1 | 14.5 | 19.0 | 22.8 |
| 1.3 | 0.7 | 8.9 | 11.9 | 10.9 | 7.3 | 5.9 | 3.7 | 17.9 | 19.2 | 9.8 |
| 5.8 | 6.9 | 2.6 | 5.8 | 21.7 | 11.8 | 3.4 | 2.1 | 4.5 | 6.3 | 10.7 |
| 8.9 | 9.4 | 9.4 | 7.6 | 10.0 | 3.3 | 6.7 | 7.8 | 11.6 | 13.8 | 18.6 |
The sample mean = 11.49 and the sample standard deviation = 6.23.
We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.
Let X = length, in seconds, of an eight-week-old baby's smile.
The notation for the uniform distribution is
X ~ U(a, b) where a = the lowest value of x and b = the highest value of x.
The probability density function is f(x) = for a ≤ x ≤ b.
For this example, X ~ U(0, 23) and f(x) = for 0 ≤ X ≤ 23.
Formulas for the theoretical mean and standard deviation are
and
For this problem, the theoretical mean and standard deviation are
μ = = 11.50 seconds and σ = = 6.64 seconds.
Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.
The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and b. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.
| 1 | 12 | 4 | 10 | 4 | 14 | 11 |
| 7 | 11 | 4 | 13 | 2 | 4 | 6 |
| 3 | 10 | 0 | 12 | 6 | 9 | 10 |
| 5 | 13 | 4 | 10 | 14 | 12 | 11 |
| 6 | 10 | 11 | 0 | 11 | 13 | 2 |
a is zero; b is 14; X ~ U (0, 14); μ = 7 passengers; σ = 4.04 passengers
a. Refer to Example 5.1. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?
b. Find the 90th percentile for an eight-week-old baby's smiling time.
b. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90
c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS.
c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds.
Find P(x > 12|x > 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds.
Write a new f(x): f(x) = =
for 8 < x < 23
P(x > 12|x > 8) = (23 − 12) =
For the second way, use the conditional formula from Probability Topics with the original distribution X ~ U (0, 23):
P(A|B) =
For this problem, A is (x > 12) and B is (x > 8).
So, P(x > 12|x > 8) =
A distribution is given as X ~ U (0, 20). What is P(2 < x < 18)? Find the 90th percentile.
P(2 < x < 18) = 0.8; 90th percentile = 18
The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.
a. What is the probability that a person waits fewer than 12.5 minutes?
a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = = for 0 ≤ x ≤ 15.
Find P (x < 12.5). Draw a graph.
The probability a person waits less than 12.5 minutes is 0.8333.
b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.
b. μ = = = 7.5. On the average, a person must wait 7.5 minutes.
c. Ninety percent of the time, the time a person must wait falls below what value?
c. Find the 90th percentile. Draw a graph. Let k = the 90th percentile.
The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.
Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X = the time, in minutes, it takes a nine-year old child to eat a donut. Then X ~ U (0.5, 4).
a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.
b. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.
The second question has a conditional probability. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.1). You must reduce the sample space. First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.
Write a new f(x):
f(x) = = for 1.5 ≤ x ≤ 4.
Find P(x > 2|x > 1.5). Draw a graph.
P(x > 2|x > 1.5) = (base)(new height) = (4 − 2)= ?
b.
The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is .
Second way: Draw the original graph for X ~ U (0.5, 4). Use the conditional formula
P(x > 2|x > 1.5) =
Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. Then X ~ U (6, 15).
Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.
P (x > 8) = 0.7778
P (x > 8 | x > 7) = 0.875
Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x = the time needed to fix a furnace. Then x ~ U (1.5, 4).
a. To find f(x): f (x) = = so f(x) = 0.4
P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8
b. P(x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6
The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0; since X ~ U (1.5, 4), x can not be less than 1.5.
c.
d.
The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car.
The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.
The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.
Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.
X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m, the decay parameter.
. Therefore,
The standard deviation, σ, is the same as the mean. μ = σ
The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25).
The probability density function is f(x) = me-mx. The number e = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key "ex." If you enter one for x, the calculator will display the value e.
The curve is:
f(x) = 0.25e–0.25x where x is at least zero and m = 0.25.
For example, f(5) = 0.25e−(0.25)(5) = 0.072. The postal clerk spends five minutes with the customers.
The graph is as follows:
Notice the graph is a declining curve. When x = 0,
f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m. The maximum value on the y-axis is m.
The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
X ~ Exp(0.125); f(x) = 0.125e–0.125x;
a. Using the information in Try It, find the probability that a clerk spends four to five minutes with a randomly selected customer.
a. Find P(4 < x < 5).
You can do these calculations easily on a calculator.
On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).
b. Half of all customers are finished within how long? (Find the 50th percentile)
b. Find the 50th percentile.
P(x < k) = 0.50, k = 2.8 minutes (calculator or computer)
Half of all customers are finished within 2.8 minutes.
You can also do the calculation as follows:
P(x < k) = 0.50 and P(x < k) = 1 –e–0.25k
Therefore, 0.50 = 1 − e−0.25k and e−0.25k = 1 − 0.50 = 0.5
Take natural logs: ln(e–0.25k) = ln(0.50). So, –0.25k = ln(0.50)
Solve for k: minutes. The calculator simplifies the calculation for percentile k. See the following two notes.
A formula for the percentile k is where ln is the natural log.
On the home screen, enter ln(1 – 0.50)/–0.25. Press the (-) for the negative.
c. Which is larger, the mean or the median?
c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.
The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?
P(x < 10) = 0.4866
50th percentile = 10.40
Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.
Let X = the amount of money a student in your class has in his or her pocket or purse.
The distribution for X is approximately exponential with mean, μ = _______ and m = _______. The standard deviation, σ = ________.
Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $.40 in his or her pocket or purse. (Shade P(x < 0.40)).
On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.
a. What is the probability that a computer part lasts more than 7 years?
a. Let x = the amount of time (in years) a computer part lasts.
On the home screen, enter e^(-.1*7).
b. On the average, how long would five computer parts last if they are used one after another?
b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.
c. Eighty percent of computer parts last at most how long?
c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.
Solve for k: years
Eighty percent of the computer parts last at most 16.1 years.
On the home screen, enter
d. What is the probability that a computer part lasts between nine and 11 years?
d. Find P(9 < x < 11). Draw the graph.
P(9 < x < 11) = P(x < 11) – P(x < 9) = (1 – e(–0.1)(11)) – (1 – e(–0.1)(9)) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737.
On the home screen, enter e^(–0.1*9) – e^(–0.1*11).
On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?
P(x > 15) = 0.4346
Six pairs of running shoes would last 108 months on average.
80th percentile = 28.97 months
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = . If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes.
What is m, μ, and σ? The probability that you must wait more than five minutes is _______ .
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter . Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What is the probability that a person is willing to commute more than 25 miles?
m = ; μ = 20; σ = 20; P(x > 25) = 0.2865
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.
1 - e^(–0.5) ≈ 0.3935
1 – (1 – e^( – 5*0.5)) or e^( – 5*0.5)
Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.
In Example 5.6 recall that the amount of time between customers is exponentially distributed with a mean of two minutes (X ~ Exp (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says that
P (X > r + t | X > r) = P (X > t) for all r ≥ 0 and t ≥ 0
For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation.
P(X > 5 + 1 | X > 5) = P(X > 1) = ≈ 0.6065.
This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example 5.8, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P(X > 17|X > 10) = P(X > 7) = 0.4966.
Refer to Example 5.6 where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?
The decay parameter of X is m = = 0.25, so X ∼ Exp(0.25).
The cumulative distribution function is P(X < x) = 1 – e–0.25x.
We want to find P(X > 7|X > 4). The memoryless property says that P(X > 7|X > 4) = P (X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.
This is P(X > 3) = 1 – P (X < 3) = 1 – (1 – e–0.25⋅3) = e–0.75 ≈ 0.4724.
1–(1–e^(–0.25*2)) = e^(–0.25*2).
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.
Let T = the lifetime of the light bulb. Then T ∼ Exp.
The cumulative distribution function is P (T < t) = 1 −
We need to find P(T > 19|T = 12). By the memoryless property,
P(T>19|T = 12) = P(T > 7) = 1 – P(T < 7) = 1 – (1 – e–7/8)= e-7/8 ≈ 0.4169.
1 – (1 – e^(–7/8)) = e^(–7/8).
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ, then . Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. (k! = k*(k-1*)(k–2)*(k-3)…3*2*1)
Suppose X has the Poisson distribution with mean λ. Compute P(X = k) by entering 2nd, VARS(DISTR), C: poissonpdf(λ, k). To compute P(X ≤ k), enter 2nd, VARS (DISTR), D:poissoncdf(λ, k).
At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.
poissonpdf(4, 5) = 0.1563.
poisssoncdf(4, 4) = 0.6288
1 – poissoncdf(32, 40). = 0.0707
In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.
poissoncdf(3, 2)
e^(-3*2).
Class Time:
Names:
Collect the DataUse a random number generator to generate 50 values between zero and one (inclusive). List them in Table 5.3. Round the numbers to four decimal places or set the calculator MODE to four places.
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a < x < b). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f(x), is used to draw the graph of the probability distribution. The total area under the graph of f(x) is one. The area under the graph of f(x) and between values a and b gives the probability P(a < x < b).
The cumulative distribution function (cdf) of X is defined by P (X ≤ x). It is a function of x that gives the probability that the random variable is less than or equal to x.
If X has a uniform distribution where a < x < b or a ≤ x ≤ b, then X takes on values between a and b (may include a and b). All values x are equally likely. We write X ∼ U(a, b). The mean of X is . The standard deviation of X is . The probability density function of X is for a ≤ x ≤ b. The cumulative distribution function of X is P(X ≤ x) = . X is continuous.
The probability P(c < X < d) may be found by computing the area under f(x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.
If X has an exponential distribution with mean μ, then the decay parameter is m = , and we write X ∼ Exp(m) where x ≥ 0 and m > 0 . The probability density function of X is f(x) = me-mx (or equivalently . The cumulative distribution function of X is P(X ≤ x) = 1 – e–mx.
The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that P(X > x + k|X > x) = P(X > k).
If T represents the waiting time between events, and if T ∼ Exp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of PX is . This may be computed using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf(λ, k). The cumulative distribution function P(X ≤ k) may be computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf(λ, k).
Probability density function (pdf) f(x):
Cumulative distribution function (cdf): P(X ≤ x)
X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X; b = largest X
X ~ U (a, b)
The mean is
The standard deviation is
Probability density function: for
Area to the Left of x: P(X < x) = (x – a)
Area to the Right of x: P(X > x) = (b – x)
Area Between c and d: P(c < x < d) = (base)(height) = (d – c)
Uniform: X ~ U(a, b) where a < x < b
Exponential: X ~ Exp(m) where m = the decay parameter
Which type of distribution does the graph illustrate?
Which type of distribution does the graph illustrate?
Which type of distribution does the graph illustrate?
What does the shaded area represent? P(___< x < ___)
What does the shaded area represent? P(___< x < ___)
For a continuous probablity distribution, 0 ≤ x ≤ 15. What is P(x > 15)?
What is the area under f(x) if the function is a continuous probability density function?
For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)?
A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)?
f(x) for a continuous probability function is , and the function is restricted to 0 ≤ x ≤ 5. What is P(x < 0)?
f(x), a continuous probability function, is equal to , and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)?
Find the probability that x falls in the shaded area.
Find the probability that x falls in the shaded area.
Find the probability that x falls in the shaded area.
f(x), a continuous probability function, is equal to and the function is restricted to 1 ≤ x ≤ 4. Describe
Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.
| 1.5 | 2.4 | 3.6 | 2.6 | 1.6 | 2.4 | 2.0 |
| 3.5 | 2.5 | 1.8 | 2.4 | 2.5 | 3.5 | 4.0 |
| 2.6 | 1.6 | 2.2 | 1.8 | 3.8 | 2.5 | 1.5 |
| 2.8 | 1.8 | 4.5 | 1.9 | 1.9 | 3.1 | 1.6 |
The sample mean = 2.50 and the sample standard deviation = 0.8302.
The distribution can be written as X ~ U(1.5, 4.5).
What type of distribution is this?
In this distribution, outcomes are equally likely. What does this mean?
What is the height of f(x) for the continuous probability distribution?
What are the constraints for the values of x?
Graph P(2 < x < 3).
What is P(2 < x < 3)?
What is P(x < 3.5| x < 4)?
What is P(x = 1.5)?
What is the 90th percentile of square footage for homes?
Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.
What is a? What does it represent?
What is b? What does it represent?
What is the probability density function?
What is the theoretical mean?
What is the theoretical standard deviation?
Draw the graph of the distribution for P(x > 9).
Find P(x > 9).
Find the 40th percentile.
What is being measured here?
In words, define the random variable X.
Are the data discrete or continuous?
The interval of values for x is ______.
The distribution for X is ______.
Write the probability density function.
Graph the probability distribution.
Find the average age of the cars in the lot.
Find the probability that a randomly chosen car in the lot was less than four years old.
Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.
What has changed in the previous two problems that made the solutions different?
Find the third quartile of ages of cars in the lot. This means you will have to find the value such that , or 75%, of the cars are at most (less than or equal to) that age.
Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)
What type of distribution is this?
Are outcomes equally likely in this distribution? Why or why not?
What is m? What does it represent?
What is the mean?
What is the standard deviation?
State the probability density function.
Graph the distribution.
Find P(2 < x < 10).
Find P(x > 6).
Find the 70th percentile.
What is m?
What is the probability density function?
What is the cumulative distribution function?
Draw the distribution.
Find P(x < 4).
Find the 30th percentile.
Find the median.
Which is larger, the mean or the median?
Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.
What is being measured here?
Are the data discrete or continuous?
In words, define the random variable X.
What is the decay rate (m)?
The distribution for X is ______.
Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P(x < 5,730).
Find the percentage of carbon-14 lasting longer than 10,000 years.
Thirty percent (30%) of carbon-14 will decay within how many years?
For each probability and percentile problem, draw the picture.
Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor.
When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why?
For each probability and percentile problem, draw the picture.
Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks).
A random number generator picks a number from one to nine in a uniform manner.
According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month.
A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution.
The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class.
Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.
What is the average waiting time (in minutes)?
Find the 30th percentile for the waiting times (in minutes).
The probability of waiting more than seven minutes given a person has waited more than four minutes is?
The time (in minutes) until the next bus departs a major bus depot follows a distribution with f(x) = where x goes from 25 to 45 minutes.
Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution.
A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution.
The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution.
Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes.
Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery.
The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state.
The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement.
The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150.
The decay rate is:
What is the probability that a phone will fail within two years of the date of purchase?
What is the median lifetime of these phones (in years)?
Let X ~ Exp(0.1).
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years.
At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution.
In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential.
During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential.
According to the American Red Cross, about one out of nine people in the U.S. have Type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B blood types that arrive roughly follows the Poisson distribution.
A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution.
At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.
McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.
Data from the United States Census Bureau.
Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013).
“No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013).
Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf (accessed June 11, 2013).
Uniform Distribution
Normal Distribution
P(6 < x < 7)
one
zero
one
0.625
The probability is equal to the area from x = to x = 4 above the x-axis and up to f(x) = .
It means that the value of x is just as likely to be any number between 1.5 and 4.5.
1.5 ≤ x ≤ 4.5
0.3333
zero
0.6
b is 12, and it represents the highest value of x.
six
4.8
X = The age (in years) of cars in the staff parking lot
0.5 to 9.5
f(x) = where x is between 0.5 and 9.5, inclusive.
μ = 5
No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.
five
f(x) = 0.2e-0.2x
0.5350
6.02
f(x) = 0.75e-0.75x
0.4756
The mean is larger. The mean is , which is greater than 0.9242.
continuous
m = 0.000121
Age is a measurement, regardless of the accuracy used.
d
b
a
c
Let T = the life time of a light bulb.
The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is
Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3.
You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is season. For the exponential, µ = .
We could also deduce that each person arriving has a 8/9 chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is . (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)
Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = . The cdf is P(T < t) =
By the end of this chapter, the student should be able to:
The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world.
In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them.
The normal distribution has two parameters (two numerical descriptive measures), the mean (μ) and the standard deviation (σ). If X is a quantity to be measured that has a normal distribution with mean (μ) and standard deviation (σ), we designate this by writing
The probability density function is a rather complicated function. Do not memorize it. It is not necessary.
f(x) =
The cumulative distribution function is P(X < x). It is calculated either by a calculator or a computer, or it is looked up in a table. Technology has made the tables virtually obsolete. For that reason, as well as the fact that there are various table formats, we are not including table instructions.
The curve is symmetrical about a vertical line drawn through the mean, μ. In theory, the mean is the same as the median, because the graph is symmetric about μ. As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Since the area under the curve must equal one, a change in the standard deviation, σ, causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ. A change in μ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution.
Your instructor will record the heights of both men and women in your class, separately. Draw histograms of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do you think that if you had recorded 200 data values for men and 200 for women that the curves would look bell-shaped? Calculate the mean for each data set. Write the means on the x-axis of the appropriate graph below the peak. Shade the approximate area that represents the probability that one randomly chosen male is taller than 72 inches. Shade the approximate area that represents the probability that one randomly chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability appear to be more than 0.5?
The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
x = μ + (z)(σ) = 5 + (3)(2) = 11
The z-score is three.
The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean μ and standard deviation σ.
If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is:
The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.
Suppose X ~ N(5, 6). This says that x is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:
This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5. The standard deviation is σ = 6.
Notice that: 5 + (2)(6) = 17 (The pattern is μ + zσ = x)
Now suppose x = 1. Then: z = = = –0.67 (rounded to two decimal places)
This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)
Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative x is less than μ.
What is the z-score of x, when x = 1 and X ~ N(12,3)?
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.
a. Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.
b. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.
b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.
Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z?
z = = = 2 where µ = 2 and σ = 1.
The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own) standard deviations to the right of their respective means.
The z-score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.
Fill in the blanks.
Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16,4). Suppose Jerome scores ten points in a game. The z–score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).
1.5, left, 16
The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule says the following:
The empirical rule is also known as the 68-95-99.7 rule.
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
b. 177.98, 1.27, right
Use the information in Example 6.2 to answer the following questions.
Solve the equation z = for x. x = μ + (z)(σ)
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males from 1984 to 1985. Then Y ~ N(172.36, 6.34).
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm?
The z-score for x = 160.58 is z = –1.5.
In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N(496, 114).
Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21?
The z-score for x1 = 325 is z1 = –1.14.
The z-score for x2 = 366.21 is z2 = –1.14.
Student 2 scored closer to the mean than Student 1 and, since they both had negative z-scores, Student 2 had the better score.
Suppose x has a normal distribution with mean 50 and standard deviation 6.
Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie?
between 20 and 30.
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N(172.36, 6.34).
The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.
The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).
The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.
Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.
To calculate the probability, use the probability tables provided in Table 13.45 without the use of technology. The tables include instructions for how to use them.
If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.
If the area to the left of x is 0.012, then what is the area to the right?
1 − 0.012 = 0.988
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
a. Find the probability that a randomly selected student scored more than 65 on the exam.
a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5
Draw a graph.
Then, find P(x > 65).
P(x > 65) = 0.3446
The probability that any student selected at random scores more than 65 is 0.3446.
Go into 2nd DISTR.
2nd DISTR, press 2:normalcdf.The syntax for the instructions are as follows:
normalcdf(lower value, upper value, mean, standard deviation)
For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter 10^99 instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve.
The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right.
z = = 0.4
Area to the left is 0.6554.
P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446
Calculate the z-score:
*Press 2nd Distr
3:invNorm(
ENTER.
2nd Distr
3:invNorm(.6554) ENTER
b. Find the probability that a randomly selected student scored less than 85.
b. Draw a graph.
Then find P(x < 85), and shade the graph.
Using a computer or calculator, find P(x < 85) = 1.
normalcdf(0,85,63,5) = 1 (rounds to one)
The probability that one student scores less than 85 is approximately one (or 100%).
c. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile.
Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.
k = 69.4
The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:
invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)
d. Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).
d. Find the 70th percentile.
Draw a new graph and label it appropriately. k = 65.6
The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.
invNorm(0.70,63,5) = 65.6
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.
Find the probability that a randomly selected golfer scored less than 65.
normalcdf(1099,65,68,3) = 0.1587
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.
Find P(1.8 < x < 2.75).
The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886
normalcdf(1.8,2.75,2,0.5) = 0.5886
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25.
invNorm(0.25,2,0.5) = 1.66
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.
normalcdf(66,70,68,3) = 0.4950
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.
c.
Use the information in Example 6.9 to answer the following questions.
Let X = a smart phone user whose age is 13 to 55+. X ~ N(36.9, 13.9)
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).
a. Calculate the interquartile range (IQR).
a.
b. Forty percent of the ages that range from 13 to 55+ are at least what age?
b.
Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points.
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
b.
c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.
c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.
Using the information from Example 6.11, answer the following:
The middle area = 0.40, so each tail has an area of 0.30.
1 – 0.40 = 0.60
The tails of the graph of the normal distribution each have an area of 0.30.
Find k1, the 30th percentile and k2, the 70th percentile (0.40 + 0.30 = 0.70).
k1 = invNorm(0.30,5.85,0.24) = 5.72 cm
k2 = invNorm(0.70,5.85,0.24) = 5.98 cm
Class Time:
Names:
DirectionsRound the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places.
| _______ | _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ | _______ |
Analyze the Distribution Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data?
Describe the Data Use the data you collected to complete the following statements.
Theoretical Distribution Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data.
Discussion QuestionsDo the data from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution, explain why or why not.
Class Time:
Names:
Collect the Data Measure the length of your pinky finger (in centimeters).
| _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ |
| _______ | _______ | _______ | _______ | _______ |
Analyze the Distribution Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution of the data you collected?
Describe the Data Using the data you collected complete the following statements. (Hint: order the data)
(IQR = Q3 – Q1)
Theoretical Distribution Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation.
Discussion QuestionsDo the data you collected give a close approximation to the theoretical distribution? In complete sentences and comparing the results in the sections titled Describe the Data and Theoretical Distribution, explain why or why not.
A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zero and the standard deviation is one. If z is the z-score for a value x from the normal distribution N(µ, σ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ.
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
X ∼ N(μ, σ)
μ = the mean σ = the standard deviation
Z ~ N(0, 1)
z = a standardized value (z-score)
mean = 0; standard deviation = 1
To find the Kth percentile of X when the z-scores is known:
z-score: z =
Z = the random variable for z-scores
Z ~ N(0, 1)
Normal Distribution: X ~ N(µ, σ) where µ is the mean and σ is the standard deviation.
Standard Normal Distribution: Z ~ N(0, 1).
Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)
Calculator function for the kth percentile: k = invNorm (area to the left of k, mean, standard deviation)
A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________.
A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?
X ~ N(1, 2)
σ = _______
A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________.
X ~ N(–4, 1)
What is the median?
X ~ N(3, 5)
σ = _______
X ~ N(–2, 1)
μ = _______
What does a z-score measure?
What does standardizing a normal distribution do to the mean?
Is X ~ N(0, 1) a standardized normal distribution? Why or why not?
What is the z-score of x = 12, if it is two standard deviations to the right of the mean?
What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean?
What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?
What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean?
Suppose X ~ N(2, 6). What value of x has a z-score of three?
Suppose X ~ N(8, 1). What value of x has a z-score of –2.25?
Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?
Suppose X ~ N(2, 3). What value of x has a z-score of –0.67?
Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?
Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean?
Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?
Suppose X ~ N(–1, 2). What is the z-score of x = 2?
Suppose X ~ N(12, 6). What is the z-score of x = 2?
Suppose X ~ N(9, 3). What is the z-score of x = 9?
Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5?
In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.
In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean.
In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean.
In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean.
In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean.
About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?
About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution?
About what percent of x values lie between the second and third standard deviations (both sides)?
Suppose X ~ N(15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15).
Suppose X ~ N(–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3).
Suppose X ~ N(–3, 1). Between what x values does 34.14% of the data lie?
About what percent of x values lie between the mean and three standard deviations?
About what percent of x values lie between the mean and one standard deviation?
About what percent of x values lie between the first and second standard deviations from the mean (both sides)?
About what percent of x values lie betwween the first and third standard deviations(both sides)?
Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.
Define the random variable X in words. X = _______________.
X ~ _____(_____,_____)
How would you represent the area to the left of one in a probability statement?
What is the area to the right of one?
Is P(x < 1) equal to P(x ≤ 1)? Why?
How would you represent the area to the left of three in a probability statement?
What is the area to the right of three?
If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x?
If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x?
Use the following information to answer the next four exercises:
X ~ N(54, 8)
Find the probability that x > 56.
Find the probability that x < 30.
Find the 80th percentile.
Find the 60th percentile.
X ~ N(6, 2)
Find the probability that x is between three and nine.
X ~ N(–3, 4)
Find the probability that x is between one and four.
X ~ N(4, 5)
Find the maximum of x in the bottom quartile.
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
Find the probability that a CD player will last between 2.8 and six years.
Find the 70th percentile of the distribution for the time a CD player lasts.
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the median recovery time?
What is the z-score for a patient who takes ten days to recover?
The length of time to find it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?
The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences.
The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution.
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14).
Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them.
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115.
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the probability of spending more than two days in recovery?
The 90th percentile for recovery times is?
Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.
Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?
Find the probability that it takes at least eight minutes to find a parking space.
Seventy percent of the time, it takes more than how many minutes to find a parking space?
According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.
In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district.
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.
Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.
Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. Table 6.3 displays the ordered real data (in minutes):
| 0.50 | 4.25 | 5 | 6 | 7.25 |
| 1.75 | 4.25 | 5.25 | 6 | 7.25 |
| 2 | 4.25 | 5.25 | 6.25 | 7.25 |
| 2.25 | 4.25 | 5.5 | 6.25 | 7.75 |
| 2.25 | 4.5 | 5.5 | 6.5 | 8 |
| 2.5 | 4.75 | 5.5 | 6.5 | 8.25 |
| 2.75 | 4.75 | 5.75 | 6.5 | 9.5 |
| 3.25 | 4.75 | 5.75 | 6.75 | 9.5 |
| 3.75 | 5 | 6 | 6.75 | 9.75 |
| 3.75 | 5 | 6 | 6.75 | 10.75 |
Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.
Table 6.4 shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.
| 40,000 | 40,000 | 45,050 | 45,500 | 46,249 | 48,134 |
| 49,133 | 50,071 | 50,096 | 50,466 | 50,832 | 51,100 |
| 51,500 | 51,900 | 52,000 | 52,132 | 52,200 | 52,530 |
| 52,692 | 53,864 | 54,000 | 55,000 | 55,000 | 55,000 |
| 55,000 | 55,000 | 55,000 | 55,082 | 57,000 | 58,008 |
| 59,680 | 60,000 | 60,000 | 60,492 | 60,580 | 62,380 |
| 62,872 | 64,035 | 65,000 | 65,050 | 65,647 | 66,000 |
| 66,161 | 67,428 | 68,349 | 68,976 | 69,372 | 70,107 |
| 70,585 | 71,594 | 72,000 | 72,922 | 73,379 | 74,500 |
| 75,025 | 76,212 | 78,000 | 80,000 | 80,000 | 82,300 |
An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.
A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.
We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following:
A $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are
Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).
“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).
“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013).
“Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013).
Data from the San Jose Mercury News.
Data from The World Almanac and Book of Facts.
“List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).
Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).
“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
“403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).
“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).
“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
“Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013).
ounces of water in a bottle
2
–4
–2
The mean becomes zero.
z = 2
z = 2.78
x = 20
x = 6.5
x = 1
x = 1.97
z = –1.67
z ≈ –0.33
0.67, right
3.14, left
about 68%
about 4%
between –5 and –1
about 50%
about 27%
The lifetime of a Sunshine CD player measured in years.
P(x < 1)
Yes, because they are the same in a continuous distribution: P(x = 1) = 0
1 – P(x < 3) or P(x > 3)
1 – 0.543 = 0.457
0.0013
56.03
0.1186
c
Let X = an SAT math score and Y = an ACT math score.
c
d
By the end of this chapter, the student should be able to:
Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem.
The central limit theorem (clt for short) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, μ, and a known standard deviation, σ. The first alternative says that if we collect samples of size n with a "large enough n," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.
In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means and the sums tend to follow the normal distribution.
The size of the sample, n, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement.
Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times.
Each time a person rolls more than one die, he or she calculates the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, 6 on one roll.
The mean is = 3.4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means.
Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5.
Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Since the "mean" when you roll one die is just the face on the die, what distribution do these means appear to be representing?
Draw the graph for the means using two dice. Do the sample means show any kind of pattern?
Draw the graph for the means using five dice. Do you see any pattern emerging?
Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice?
As the number of dice rolled increases from one to two to five to ten, the following is happening:
You have just demonstrated the central limit theorem (clt).
The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution).
Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:
If you draw random samples of size n, then as n increases, the random variable which consists of sample means, tends to be normally distributed and
~ N.
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by, the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done.
To put it more formally, if you draw random samples of size n, the distribution of the random variable , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases.
The random variable has a different z-score associated with it from that of the random variable X. The mean is the value of in one sample.
μX is the average of both X and .
= standard deviation of and is called the standard error of the mean.
To find probabilities for means on the calculator, follow these steps.
2nd DISTR
where:
An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population.
a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean.
Let = the mean of a sample of size 25. Since μX = 90, σX = 15, and n = 25,
~ N.
Find P(85 < < 92). Draw a graph.
P(85 < < 92) = 0.6997
The probability that the sample mean is between 85 and 92 is 0.6997.
normalcdf(lower value, upper value, mean, standard error of the mean)
The parameter list is abbreviated (lower value, upper value, μ, )
normalcdf(85,92,90,) = 0.6997
b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.
b. To find the value that is two standard deviations above the expected value 90, use the formula:
value = μx + (#ofTSDEVs)
value = 90 + 2 = 96
The value that is two standard deviations above the expected value is 96.
The standard error of the mean is = = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n.
An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.
P(42 < < 50) = = 0.9797
The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.
Let X = the time, in hours, it takes to play one soccer match.
The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match.
Let = the mean time, in hours, it takes to play one soccer match.
If μX = _________, σX = __________, and n = ___________, then X ~ N(______, ______) by the central limit theorem for means.
μX = 2, σX = 0.5, n = 50, and X ~ N
Find P(1.8 < < 2.3). Draw a graph.
P(1.8 < < 2.3) = 0.9977
normalcdf = 0.9977
The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.
The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.
P(2 < < 3) = normalcdf = 1
To find percentiles for means on the calculator, follow these steps.
2nd DIStR
k = invNorm
where:
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.
normalcdf(30,E99,34,1.5) = 0.9962In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy?
You need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game.
P(29 < < 35) = normalcdf = 0.0186
You can conclude there is approximately a 1.9% chance that your game will be played by men whose mean age is between 29 and 35.
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
invNorm = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.normalcdf = 0.9293Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?
We have P(( > 16.01) = normalcdf = 0.3417. Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N((n)(μΧ), ()(σΧ)).
The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
The random variable ΣX has the following z-score associated with it:
To find probabilities for sums on the calculator, follow these steps.
2nd DISTR
normalcdf
normalcdf(lower value of the area, upper value of the area, (n)(mean), ()(standard deviation))
where:
An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.
Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.
ΣX = the sum or total of 80 values. Since μX = 90, σX = 15, and n = 80, ~ N((80)(90),
a. Find P(Σx > 7,500)
P(Σx > 7,500) = 0.0127
normalcdf(lower value, upper value, mean of sums, stdev of sums)
The parameter list is abbreviated(lower, upper, (n)(μX, (σX))
normalcdf (7500,1E99,(80)(90),(15)) = 0.0127
1E99 = 1099.
Press the EE key for E.
b. Find Σx where z = 1.5.
Σx = (n)(μX) + (z)(σΧ) = (80)(90) + (1.5)()(15) = 7,401.2
An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.
0.0040
To find percentiles for sums on the calculator, follow these steps.
2nd DIStR
where:
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.
normalcdf
(1,500, 1,800, (50)(34), (15)) = 0.7974invNorm(0.80,(50)(34),(15)) = 1,789.3In a recent study reported Oct.29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.
normalcdf
(1400,1500,(39)(35),()(10)) = 0.2723invNorm(0.90,(39)(35),()
(10)) = 1445.0The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70.
normalcdf(600,E99,(70)(8.2),(1)) = 0.0009The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.
It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.
If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.
The law of large numbers says that if you take samples of larger and larger size from any population, then the mean of the sample tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for is .) This means that the sample mean must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.
A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:
Let X = one stress score.
Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75.
Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution).
μX = = = 3
σX = = = 1.15
For problems 1. and 2., let = the mean stress score for the 75 students. Then,
∼ N where n = 75.
a. Find P( < 2). Draw the graph.
a. P( < 2) = 0
The probability that the mean stress score is less than two is about zero.
normalcdf = 0
The smallest stress score is one.
b. Find the 90th percentile for the mean of 75 stress scores. Draw a graph.
b. Let k = the 90th precentile.
Find k, where P( < k) = 0.90.
k = 3.2
The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.
invNorm = 3.2
For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3),(1.15)]
c. Find P(Σx < 200). Draw the graph.
c. The mean of the sum of 75 stress scores is (75)(3) = 225
The standard deviation of the sum of 75 stress scores is (1.15) = 9.96
P(Σx < 200) = 0
The probability that the total of 75 scores is less than 200 is about zero.
normalcdf (75,200,(75)(3),(1.15)).
The smallest total of 75 stress scores is 75, because the smallest single score is one.
d. Find the 90th percentile for the total of 75 stress scores. Draw a graph.
d. Let k = the 90th percentile.
Find k where P(Σx < k) = 0.90.
k = 237.8
The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.
invNorm(0.90,(75)(3),(1.15)) = 237.8
Use the information in Example 7.7, but use a sample size of 55 to answer the following questions.
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
X ∼ Exp. From previous chapters, we know that μ = 22 and σ = 22.
Let = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance.
~ N by the central limit theorem for sample means
Find: P( > 20)
P( > 20) = 0.79199 using normalcdf
The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99.
Find P(x > 20). Remember to use the exponential distribution for an individual: .
Using the clt to find percentilesFind the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.
Let k = the 95th percentile. Find k where P( < k) = 0.95
k = 26.0 using invNorm = 26.0
The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.
Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.
Use the information in Example 7.8, but change the sample size to 144.
In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.
normalcdf(1.75,1.85,2,0.05) = 0.0013Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.
normalcdf(120,99,114.8,13.1) = 0.0272. There is about a 3%, that the randomly selected woman will have systolics blood pressure greater than 120.normalcdf = 0.006. There is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.
normalcdf(-E99,35,30.9,1.8) = 0.9886normalcdf(50, E99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.normalcdf(1600,E99,1514.10,63) = 0.0864normalcdf(-E99,1595,1514.10,63) = 0.9005. This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1595.invNorm(0.95,30.9,1.1) = 32.7. This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average.invNorm(0.90,2008.5,72.56) = 2101.5. This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years.According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.
invNorm(0.95,69,2.8) = 73.61invNorm(0.95,69,0.28) = 69.49: Normal Approximation to the Binomial
Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution:
Recall that if X is the binomial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = . Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example.
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
Let X = the number that favor a charter school for grades K trough 5. X ~ B(n, p) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = . The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447). See The Normal Distribution for help with calculator instructions.
For part a, you include 150 so P(X ≥ 150) has normal approximation P(Y ≥ 149.5) = 0.8641.
normalcdf(149.5,10^99,159,8.6447) = 0.8641.
For part b, you include 160 so P(X ≤ 160) has normal appraximation P(Y ≤ 160.5) = 0.5689.
normalcdf(0,160.5,159,8.6447) = 0.5689
For part c, you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.
normalcdf(155.5,10^99,159,8.6447) = 0.6572.
For part d, you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741.
normalcdf(0,146.5,159,8.6447) = 0.0741
For part e,P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083.
normalcdf(174.5,175.5,159,8.6447) = 0.0083
Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in "binomial probability distribution calculation" in an Internet browser, you can find at least one online calculator for the binomial.
For Example 7.9, the probabilities are calculated using the following binomial distribution: (n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial.
P(X ≥ 150) :1 - binomialcdf(300,0.53,149) = 0.8641
P(X ≤ 160) :binomialcdf(300,0.53,160) = 0.5684
P(X > 155) :1 - binomialcdf(300,0.53,155) = 0.6576
P(X < 147) :binomialcdf(300,0.53,146) = 0.0742
P(X = 175) :(You use the binomial pdf.)binomialpdf(300,0.53,175) = 0.0083
In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.
0.0401
Class Time:
Names:
This lab works best when sampling from several classes and combining data.
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
Collecting Averages of PairsRepeat steps one through five of the section Collect the Data. with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs.
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
Collecting Averages of Groups of FiveRepeat steps one through five (of the section titled Collect the Data) with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of five.
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
Class Time:
Names:
GivenX = length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of the different recipes makes the same quantity of cookies.)
| Recipe # | X | Recipe # | X | Recipe # | X | Recipe # | X | |||
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 16 | 2 | 31 | 3 | 46 | 2 | |||
| 2 | 5 | 17 | 2 | 32 | 4 | 47 | 2 | |||
| 3 | 2 | 18 | 4 | 33 | 5 | 48 | 11 | |||
| 4 | 5 | 19 | 6 | 34 | 6 | 49 | 5 | |||
| 5 | 6 | 20 | 1 | 35 | 6 | 50 | 5 | |||
| 6 | 1 | 21 | 6 | 36 | 1 | 51 | 4 | |||
| 7 | 2 | 22 | 5 | 37 | 1 | 52 | 6 | |||
| 8 | 6 | 23 | 2 | 38 | 2 | 53 | 5 | |||
| 9 | 5 | 24 | 5 | 39 | 1 | 54 | 1 | |||
| 10 | 2 | 25 | 1 | 40 | 6 | 55 | 1 | |||
| 11 | 5 | 26 | 6 | 41 | 1 | 56 | 2 | |||
| 12 | 1 | 27 | 4 | 42 | 6 | 57 | 4 | |||
| 13 | 1 | 28 | 1 | 43 | 2 | 58 | 3 | |||
| 14 | 3 | 29 | 6 | 44 | 6 | 59 | 6 | |||
| 15 | 2 | 30 | 2 | 45 | 2 | 60 | 5 |
Calculate the following:
Collect the DataUse a random number generator to randomly select four samples of size n = 5 from the given population. Record your samples in Table 7.5. Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class.
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample means from other groups: | |
|---|---|---|---|---|---|
| Means: | = ____ | = ____ | = ____ | = ____ |
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample means from other groups | |
|---|---|---|---|---|---|
| Means: | = ____ | = ____ | = ____ | = ____ |
In a population whose distribution may be known or unknown, if the size (n) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n).
The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of μX and a standard deviation of σx, the mean of the sums is nμx and the standard deviation is (σx) where n is the sample size.
The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean gets to μ.
The Central Limit Theorem for Sample Means: ~ N
The Mean : μx
Central Limit Theorem for Sample Means z-score and standard error of the mean:
Standard Error of the Mean (Standard Deviation ()):
The Central Limit Theorem for Sums: ∑X ~ N[(n)(μx),()(σx)]
Mean for Sums (∑X): (n)(μx)
The Central Limit Theorem for Sums z-score and standard deviation for sums:
Standard deviation for Sums (∑X): (σx)
Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
What is the mean, standard deviation, and sample size?
Complete the distributions.
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
What causes the probabilities in Exercise 7.3 and Exercise 7.4 to be different?
Find the 95th percentile for the mean time to complete one month's reviews. Sketch the graph.
Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.
Find the probability that the sum of the 95 values is greater than 7,650.
Find the probability that the sum of the 95 values is less than 7,400.
Find the sum that is two standard deviations above the mean of the sums.
Find the sum that is 1.5 standard deviations below the mean of the sums.
Find the probability that the sum of the 40 values is greater than 7,500.
Find the probability that the sum of the 40 values is less than 7,000.
Find the sum that is one standard deviation above the mean of the sums.
Find the sum that is 1.5 standard deviations below the mean of the sums.
Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.
Find the probability that the sum of the 100 values is greater than 3,910.
Find the probability that the sum of the 100 values is less than 3,900.
Find the probability that the sum of the 100 values falls between the numbers you found in Exercise 7.16 and Exercise 7.17.
Find the sum with a z–score of –2.5.
Find the sum with a z–score of 0.5.
Find the probability that the sums will fall between the z-scores –2 and 1.
What is the mean of ΣX?
What is the standard deviation of ΣX?
What is P(Σx = 290)?
What is P(Σx > 290)?
True or False: only the sums of normal distributions are also normal distributions.
In order for the sums of a distribution to approach a normal distribution, what must be true?
What three things must you know about a distribution to find the probability of sums?
An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42?
An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200?
What is the z-score for Σx = 840?
What is the z-score for Σx = 1,186?
What is P(Σx < 1,186)?
What is the mean of ΣX?
What is the standard deviation of ΣX?
What is P(Σx > 9,000)?
Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.
Draw the graph from Exercise 7.37
Find the probability that the mean actual weight for the 100 weights is greater than 25.2.
Draw the graph from Exercise 7.39
Find the 90th percentile for the mean weight for the 100 weights.
Draw the graph from Exercise 7.41
Draw the graph from Exercise 7.43
Find the 90th percentile for the total weight of the 100 weights.
Draw the graph from Exercise 7.45
What is the distribution for the length of time one battery lasts?
What is the distribution for the mean length of time 64 batteries last?
What is the distribution for the total length of time 64 batteries last?
Find the probability that the sample mean is between seven and 11.
Find the 80th percentile for the total length of time 64 batteries last.
Find the IQR for the mean amount of time 64 batteries last.
Find the middle 80% for the total amount of time 64 batteries last.
Find P(Σx > 420).
Find the 90th percentile for the sums.
Find the 15th percentile for the sums.
Find the first quartile for the sums.
Find the third quartile for the sums.
Find the 80th percentile for the sums.
Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.
According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.
Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let the average of the 49 races.
The length of songs in a collector’s iTunes album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.
In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.
Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let = average percent of fat calories.
The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.
Which of the following is NOT TRUE about the distribution for averages?
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:
Which of the following is NOT TRUE about the theoretical distribution of sums?
Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.
Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.
Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district.
The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.
The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows.
8.625 30.25 27.625 46.75 32.875 18.25 5 0.125 2.9375 6.875 28.25 24.25 21 1.5 30.25 71 43.5 49.25 2.5625 31 16.5 9.5 18.5 18 9 10.5 16.625 1.25 18 12.87 7 12.875 2.875 60.25 29.25
Χ ~ _____(_____,_____)
The average wait time is:
Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is:
Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited.
The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is:
Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?
What's the approximate probability that the average price for 16 gas stations is over $4.69?
Find the probability that the average price for 30 gas stations is less than $4.55.
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate following using the normal approximation to the binomial distribtion.
If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology.
Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places.
X ~ N(60, 9). Suppose that you form random samples of 25 from this distribution. Let be the random variable of averages. Let ΣX be the random variable of sums. For parts c through f, sketch the graph, shade the region, label and scale the horizontal axis for , and find the probability.
Suppose that the length of research papers is uniformly distributed from ten to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers.
Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district.
The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.
NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company’s claim reasonable?
Men have an average weight of 172 pounds with a standard deviation of 29 pounds.
M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class.
| Red | Orange | Yellow | Brown | Blue | Green |
|---|---|---|---|---|---|
| 0.751 | 0.735 | 0.883 | 0.696 | 0.881 | 0.925 |
| 0.841 | 0.895 | 0.769 | 0.876 | 0.863 | 0.914 |
| 0.856 | 0.865 | 0.859 | 0.855 | 0.775 | 0.881 |
| 0.799 | 0.864 | 0.784 | 0.806 | 0.854 | 0.865 |
| 0.966 | 0.852 | 0.824 | 0.840 | 0.810 | 0.865 |
| 0.859 | 0.866 | 0.858 | 0.868 | 0.858 | 1.015 |
| 0.857 | 0.859 | 0.848 | 0.859 | 0.818 | 0.876 |
| 0.942 | 0.838 | 0.851 | 0.982 | 0.868 | 0.809 |
| 0.873 | 0.863 | 0.803 | 0.865 | ||
| 0.809 | 0.888 | 0.932 | 0.848 | ||
| 0.890 | 0.925 | 0.842 | 0.940 | ||
| 0.878 | 0.793 | 0.832 | 0.833 | ||
| 0.905 | 0.977 | 0.807 | 0.845 | ||
| 0.850 | 0.841 | 0.852 | |||
| 0.830 | 0.932 | 0.778 | |||
| 0.856 | 0.833 | 0.814 | |||
| 0.842 | 0.881 | 0.791 | |||
| 0.778 | 0.818 | 0.810 | |||
| 0.786 | 0.864 | 0.881 | |||
| 0.853 | 0.825 | ||||
| 0.864 | 0.855 | ||||
| 0.873 | 0.942 | ||||
| 0.880 | 0.825 | ||||
| 0.882 | 0.869 | ||||
| 0.931 | 0.912 | ||||
| 0.887 |
The bag contained 465 candies and he listed weights in the table came from randomly selected candies. Count the weights.
The Screw Right Company claims their inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded.
| 0.757 | 0.723 | 0.754 | 0.737 | 0.757 | 0.741 | 0.722 | 0.741 | 0.743 | 0.742 |
| 0.740 | 0.758 | 0.724 | 0.739 | 0.736 | 0.735 | 0.760 | 0.750 | 0.759 | 0.754 |
| 0.744 | 0.758 | 0.765 | 0.756 | 0.738 | 0.742 | 0.758 | 0.757 | 0.724 | 0.757 |
| 0.744 | 0.738 | 0.763 | 0.756 | 0.760 | 0.768 | 0.761 | 0.742 | 0.734 | 0.754 |
| 0.758 | 0.735 | 0.740 | 0.743 | 0.737 | 0.737 | 0.725 | 0.761 | 0.758 | 0.756 |
The screws were randomly selected from the local home repair store.
Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ tesst, what is the probability that the sample mean scores will be between 85 and 125 points?
Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?
Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013).
Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013).
Data from the United States Department of Agriculture.
Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May 17, 2013).
Data from the Wall Street Journal.
“National Health and Nutrition Examination Survey.” Center for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013).
mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
The fact that the two distributions are different accounts for the different probabilities.
0.3345
7,833.92
0.0089
7,326.49
77.45%
0.4207
3,888.5
0.8186
5
0.9772
The sample size, n, gets larger.
49
26.00
0.1587
1,000
0.0003
25.07
2,507.40
N
0.7799
1.69
0.0072
391.54
405.51
b
b
b
a
normalcdf(lower, upper, μ, ) = normalcdf = 0.0200.normalcdf(396.9,E99,(465)(0.8565),(0.05)()) ≈ 1Use normalcdf = 0.7986. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours.
Since we have normalcdf ≈ 1, we can conclude that practically all the coins are within the limits, therefore, there should be no rejected coins out of a well selected sample of size 280.
By the end of this chapter, the student should be able to:
Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion.
We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics. The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals.
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from iTunes. If so, you could conduct a survey and calculate the sample mean, , and the sample standard deviation, s. You would use to estimate the population mean and s to estimate the population standard deviation. The sample mean, , is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ.
Each of and s is called a statistic.
A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter.
Suppose, for the iTunes example, we do not know the population mean μ, but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is
.
The empirical rule, which applies to bell-shaped distributions, says that in approximately 95% of the samples, the sample mean, , will be within two standard deviations of the population mean μ. For our iTunes example, two standard deviations is (2)(0.1) = 0.2. The sample mean is likely to be within 0.2 units of μ.
Because is within 0.2 units of μ, which is unknown, then μ is likely to be within 0.2 units of in 95% of the samples. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between and in 95% of all the samples.
For the iTunes example, suppose that a sample produced a sample mean . Then the unknown population mean μ is between
and
We say that we are 95% confident that the unknown population mean number of songs downloaded from iTunes per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2).
The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ or our sample produced an that is not within 0.2 units of the true mean μ. The second possibility happens for only 5% of all the samples (95–100%).
Remember that a confidence interval is created for an unknown population parameter like the population mean, μ. Confidence intervals for some parameters have the form:
(point estimate – margin of error, point estimate + margin of error)
The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean.
When you read newspapers and journals, some reports will use the phrase "margin of error." Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. These are two ways of expressing the same concept.
Although the text only covers symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation).
Have your instructor record the number of meals each student in your class eats out in a week. Assume that the standard deviation is known to be three meals. Construct an approximate 95% confidence interval for the true mean number of meals students eat out each week.
We say we are approximately 95% confident that the true mean number of meals that students eat out in a week is between __________ and ___________.
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of and we have constructed the 90% confidence interval (5, 15) where EBM = 5.
To construct a confidence interval for a single unknown population mean μ, where the population standard deviation is known, we need as an estimate for μ and we need the margin of error. Here, the margin of error (EBM) is called the error bound for a population mean (abbreviated EBM). The sample mean is the point estimate of the unknown population mean μ.
The confidence interval estimate will have the form:
(point estimate - error bound, point estimate + error bound) or, in symbols,()
The margin of error (EBM) depends on the confidence level (abbreviated CL). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions.
There is another probability called alpha (α). α is related to the confidence level, CL. α is the probability that the interval does not contain the unknown population parameter.
= 7 and EBM = 2.5
The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5).
If the confidence level (CL) is 95%, then we say that, "We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5."
Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2.
What is the confidence interval estimate for the population mean?
(11.8, 18.2)
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of = 10, and we have constructed the 90% confidence interval (5, 15) where EBM = 5.
To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the normal distribution.
To capture the central 90%, we must go out 1.645 "standard deviations" on either side of the calculated sample mean. The value 1.645 is the z-score from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail.
It is important that the "standard deviation" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is . The fraction , is commonly called the "standard error of the mean" in order to distinguish clearly the standard deviation for a mean from the population standard deviation σ.
To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:
We will first examine each step in more detail, and then illustrate the process with some examples.
When we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1).
The confidence level, CL, is the area in the middle of the standard normal distribution. CL = 1 – α, so α is the area that is split equally between the two tails. Each of the tails contains an area equal to .
The z-score that has an area to the right of is denoted by .
For example, when CL = 0.95, α = 0.05 and = 0.025; we write = z0.025.
The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975.
, using a calculator, computer or a standard normal probability table.
invNorm(0.975, 0, 1) = 1.96
Remember to use the area to the LEFT of ; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z ~ N(0, 1).
The error bound formula for an unknown population mean μ when the population standard deviation σ is known is
The graph gives a picture of the entire situation.
CL + + = CL + α = 1.
The interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)."
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Solution ATo find the confidence interval, you need the sample mean, , and the EBM.
CL = 0.90 so α = 1 – CL = 1 – 0.90 = 0.10
= 0.05
The area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 1 – 0.05 = 0.95.
using invNorm(0.95, 0, 1) on the TI-83,83+, and 84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645) = 0.8225
- EBM = 68 - 0.8225 = 67.1775
+ EBM = 68 + 0.8225 = 68.8225
The 90% confidence interval is (67.1775, 68.8225).
Solution B
Press STAT and arrow over to TESTS.
7:ZInterval.
ENTER.
Stats and press ENTER.
C-level.
Calculate and press ENTER.
InterpretationWe estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
Explanation of 90% Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes.
Find a 90% confidence interval estimate for the population mean delivery time.
(34.1347, 37.8653)
The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table 8.1 shows the highest SAR level for a random selection of cell phone models as measured by the FCC.
| Phone Model | SAR | Phone Model | SAR | Phone Model | SAR |
|---|---|---|---|---|---|
| Apple iPhone 4S | 1.11 | LG Ally | 1.36 | Pantech Laser | 0.74 |
| BlackBerry Pearl 8120 | 1.48 | LG AX275 | 1.34 | Samsung Character | 0.5 |
| BlackBerry Tour 9630 | 1.43 | LG Cosmos | 1.18 | Samsung Epic 4G Touch | 0.4 |
| Cricket TXTM8 | 1.3 | LG CU515 | 1.3 | Samsung M240 | 0.867 |
| HP/Palm Centro | 1.09 | LG Trax CU575 | 1.26 | Samsung Messager III SCH-R750 | 0.68 |
| HTC One V | 0.455 | Motorola Q9h | 1.29 | Samsung Nexus S | 0.51 |
| HTC Touch Pro 2 | 1.41 | Motorola Razr2 V8 | 0.36 | Samsung SGH-A227 | 1.13 |
| Huawei M835 Ideos | 0.82 | Motorola Razr2 V9 | 0.52 | SGH-a107 GoPhone | 0.3 |
| Kyocera DuraPlus | 0.78 | Motorola V195s | 1.6 | Sony W350a | 1.48 |
| Kyocera K127 Marbl | 1.25 | Nokia 1680 | 1.39 | T-Mobile Concord | 1.38 |
Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. Assume that the population standard deviation is σ = 0.337.
Solution ATo find the confidence interval, start by finding the point estimate: the sample mean.
Next, find the EBM. Because you are creating a 98% confidence interval, CL = 0.98.
You need to find z0.01 having the property that the area under the normal density curve to the right of z0.01 is 0.01 and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find z0.01 = 2.326.
To find the 98% confidence interval, find .
– EBM = 1.024 – 0.1431 = 0.8809
– EBM = 1.024 – 0.1431 = 1.1671
We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram.
Solution B
Table 8.2 shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337.
| Phone Model | SAR | Phone Model | SAR |
|---|---|---|---|
| Blackberry Pearl 8120 | 1.48 | Nokia E71x | 1.53 |
| HTC Evo Design 4G | 0.8 | Nokia N75 | 0.68 |
| HTC Freestyle | 1.15 | Nokia N79 | 1.4 |
| LG Ally | 1.36 | Sagem Puma | 1.24 |
| LG Fathom | 0.77 | Samsung Fascinate | 0.57 |
| LG Optimus Vu | 0.462 | Samsung Infuse 4G | 0.2 |
| Motorola Cliq XT | 1.36 | Samsung Nexus S | 0.51 |
| Motorola Droid Pro | 1.39 | Samsung Replenish | 0.3 |
| Motorola Droid Razr M | 1.3 | Sony W518a Walkman | 0.73 |
| Nokia 7705 Twist | 0.7 | ZTE C79 | 0.869 |
Z0.035 = 1.812
– EBM = 0.940 – 0.1365 = 0.8035
+ EBM = 0.940 + 0.1365 = 1.0765
We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram.
Notice the difference in the confidence intervals calculated in Example 8.2 and the following Try It exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.
Suppose we change the original problem in Example 8.1 by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.
To find the confidence interval, you need the sample mean, , and the EBM.
CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05
The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975.
when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.)
EBM = (1.96) = 0.98
– EBM = 68 – 0.98 = 67.02
+ EBM = 68 + 0.98 = 68.98
Notice that the EBM is larger for a 95% confidence level in the original problem.
InterpretationWe estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.
Explanation of 95% Confidence Level Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score.
Comparing the results The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider.
Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.
(33.37, 38.63)
Suppose we change the original problem in Example 8.1 to see what happens to the error bound if the sample size is changed.
Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36?
Solution AIf we increase the sample size n to 100, we decrease the error bound.
When n = 100: EBM = = (1.645) = 0.4935.
Solution BIf we decrease the sample size n to 25, we increase the error bound.
When n = 25: EBM = = (1.645) = 0.987.
Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.
(34.6041, 37.3958)
When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.
Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean.
Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.
Sample mean is 45, error bound is 2.88
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population mean when the population standard deviation is known is
The formula for sample size is n = , found by solving the error bound formula for n.
In this formula, z is , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?
Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students.
The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?
35 students
In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "Student."
Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ.
If you draw a simple random sample of size n from a population that has an approximately a normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t = , then the t-scores follow a Student's t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score. It measures how far is from its mean μ. For each sample size n, there is a different Student's t-distribution.
The degrees of freedom, n – 1, come from the calculation of the sample standard deviation s. In Table 13.45, we used n deviations to calculate s. Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df).
Calculators and computers can easily calculate any Student's t-probabilities. The TI-83,83+, and 84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified.
A probability table for the Student's t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-Distribution.) When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.
If the population standard deviation is not known, the error bound for a population mean is:
The format for the confidence interval is:
To calculate the confidence interval directly:
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
Solution ATo find the confidence interval, you need the sample mean, , and the EBM.
= 8.2267 s = 1.6722 n = 15
df = 15 – 1 = 14 CL so α = 1 – CL = 1 – 0.95 = 0.05
= 0.025
The area to the right of t0.025 is 0.025, and the area to the left of t0.025 is 1 – 0.025 = 0.975
using invT(.975,14) on the TI-84+ calculator.
– EBM = 8.2267 – 0.9240 = 7.3
+ EBM = 8.2267 + 0.9240 = 9.15
The 95% confidence interval is (7.30, 9.15).
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
Press STAT and arrow over to TESTS.
8:TInterval and press ENTER (or you can just press 8).
Data and press ENTER.
List and enter the list name where you put the data.
Freq.
C-level and enter 0.95
Calculate and press ENTER.
When calculating the error bound, a probability table for the Student's t-distribution can also be used to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table.
You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5
(8.1634, 9.8032)
The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the "In utero/newborn" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 8.3 shows how many of the targeted chemicals were found in each infant’s cord blood.
| 79 | 145 | 147 | 160 | 116 | 100 | 159 | 151 | 156 | 126 |
| 137 | 83 | 156 | 94 | 121 | 144 | 123 | 114 | 139 | 99 |
Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood.
Solution AFrom the sample, you can calculate = 127.45 and s = 25.965. There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19.
You are asked to calculate a 90% confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10
By definition, the area to the right of t0.05 is 0.05 and so the area to the left of t0.05 is 1 – 0.05 = 0.95.
Use a table, calculator, or computer to find that t0.05 = 1.729.
– EBM = 127.45 – 10.038 = 117.412
+ EBM = 127.45 + 10.038 = 137.488
We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.
Solution B
Enter the data as a list.
STAT and arrow over to TESTS.
8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER.
List and enter the list name where you put the data.
Freq and enter 1.
C-level and enter 0.90
Calculate and press ENTER.
A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table 8.4. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.
| 0 | 3 | 1 | 20 | 9 |
| 5 | 10 | 1 | 10 | 4 |
| 14 | 2 | 4 | 4 | 5 |
= 6.133, s = 5.514, n = 15, and df = 15 – 1 = 14
CL = 0.98, so α = 1 - CL = 1 - 0.98 = 0.02
– EBM = 6.133 – 3.736 = 2.397
+ EBM = 6.133 + 3.736 = 9.869
We estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.
Enter the data as a list.
STAT and arrow over to TESTS.8:TInterval.ENTER.Data and press ENTER.Freq: 1C-Level: 0.98Calculate and press Enter.During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.
How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B(n, p) where n is the number of trials and p is the probability of a success. To form a proportion, take X, the random variable for the number of successes and divide it by n, the number of trials (or the sample size). The random variable P′ (read "P prime") is that proportion,
(Sometimes the random variable is denoted as , read "P hat".)
When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial.
If we divide the random variable, the mean, and the standard deviation by n, we get a normal distribution of proportions with P′, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by n.)
Using algebra to simplify :
P′ follows a normal distribution for proportions:
The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion.
p′ =
p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.)
x = the number of successes
n = the size of the sample
The error bound for a proportion is
where q′ = 1 – p′
This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is . For a proportion, the appropriate standard deviation is .
However, in the error bound formula, we use as the standard deviation, instead of .
In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q. The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures.
The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five.
For the normal distribution of proportions, the z-score formula is as follows.
If then the z-score formula is
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Let X = the number of people in the sample who have cell phones. X is binomial. .
To calculate the confidence interval, you must find p′, q′, and EBP.
n = 500
x = the number of successes = 421
p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion.
q′ = 1 – p′ = 1 – 0.842 = 0.158
Since CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 = 0.025.
Then
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.810, 0.874).
InterpretationWe estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence LevelNinety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B
Press STAT and arrow over to TESTS.
A:1-PropZint. Press ENTER.C-Level and enter .95.Calculate and press ENTER.Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
(0.3315, 0.4525)
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Solution A x = 300 and n = 500
Since CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10 = 0.05
= z0.05 = 1.645
Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z0.05. Remember that the area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.
The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.564,0.636).
Explanation of 90% Confidence LevelNinety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Solution B
Press STAT and arrow over to TESTS.
A:1-PropZint. Press ENTER.
C-Level and enter 0.90.
Calculate and press ENTER.
A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.
a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.
b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.
Sixty-eight percent (68%) of students own an iPod and a smart phone.
Since CL = 0.97, we know α = 1 – 0.97 = 0.03 and = 0.015.
The area to the left of z0.015 is 0.015, and the area to the right of z0.015 is 1 – 0.015 = 0.985.
Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1),
p′ – EPB = 0.68 – 0.0269 = 0.6531
p′ + EPB = 0.68 + 0.0269 = 0.7069
We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.
Press STAT and arrow over to TESTS.
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2.
Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke.
Solution ASix students out of 25 reported smoking within the past week, so x = 6 and n = 25. Because we are using the “plus-four” method, we will use x = 6 + 2 = 8 and n = 25 + 4 = 29.
Since CL = 0.95, we know α = 1 – 0.95 = 0.05 and = 0.025.
p′ – EPB = 0.276 – 0.163 = 0.113
p′ + EPB = 0.276 + 0.163 = 0.439
We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.
Solution B
Press STAT and arrow over to TESTS.
Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.
Arrow down to x and enter eight.
Out of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.
Using “plus four,” we have x = 31 + 2 = 33 and n = 65 + 4 = 69.
Since CL = 0.96, we know α = 1 – 0.96 = 0.04 and = 0.02.
p′ – EPB = 0.478 – 0.124 = 0.354
p′ + EPB = 0.478 + 0.124 = 0.602
We are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.
Press STAT and arrow over to TESTS.
The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.
Solution AUsing “plus-four,” we have x = 13 + 2 = 15 and n = 50 + 4 = 54.
Since CL = 0.90, we know α = 1 – 0.90 = 0.10 and = 0.05.
p′ – EPB = 0.278 – 0.100 = 0.178
p′ + EPB = 0.278 + 0.100 = 0.378
We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.
Solution B
The Berkman Center Study referenced in Example 8.12 talked to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the “plus-four” method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example 8.12.
Using “plus-four,” we have x = 159 + 2 = 161 and n = 588 + 4 = 592.
Since CL = 0.90, we know α = 1 – 0.90 = 0.10 and = 0.05
p′ – EPB = 0.272 – 0.030 = 0.242
p′ + EPB = 0.272 + 0.030 = 0.302
We are 90% confident that between 24.2% and 30.2% of all teens would report having more than 500 friends on Facebook.
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 161.
Arrow down to n and enter 592.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.242, 0.302).
Conclusion: The confidence interval for the larger sample is narrower than the interval from Example 8.12. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB, changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population proportion is
Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.
From the problem, we know that EBP = 0.03 (3%=0.03) and z0.05 = 1.645 because the confidence level is 90%.
However, in order to find n, we need to know the estimated (sample) proportion p′. Remember that q′ = 1 – p′. But, we do not know p′ yet. Since we multiply p′ and q′ together, we make them both equal to 0.5 because p′q′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n. This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size n, use the formula and make the substitutions.
gives
Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.
Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?
271 customers should be surveyed.Check the Real Estate section in your local
Class Time:
Names:
Collect the Data Check the Real Estate section in your local newspaper. Record the sale prices for 35 randomly selected homes recently listed in the county.
Many newspapers list them only one day per week. Also, we will assume that homes come up for sale randomly.
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| Confidence level | EBM/Error Bound | Confidence Interval |
|---|---|---|
| 50% | ||
| 80% | ||
| 95% | ||
| 99% |
Class Time:
Names:
| Confidence level | EBP/Error Bound | Confidence Interval |
|---|---|---|
| 50% | ||
| 80% | ||
| 95% | ||
| 99% |
Class Time:
Names:
Given:
| 59.4 | 71.6 | 69.3 | 65.0 | 62.9 | 66.5 | 61.7 | 55.2 |
| 67.5 | 67.2 | 63.8 | 62.9 | 63.0 | 63.9 | 68.7 | 65.5 |
| 61.9 | 69.6 | 58.7 | 63.4 | 61.8 | 60.6 | 69.8 | 60.0 |
| 64.9 | 66.1 | 66.8 | 60.6 | 65.6 | 63.8 | 61.3 | 59.2 |
| 64.1 | 59.3 | 64.9 | 62.4 | 63.5 | 60.9 | 63.3 | 66.3 |
| 61.5 | 64.3 | 62.9 | 60.6 | 63.8 | 58.8 | 64.9 | 65.7 |
| 62.5 | 70.9 | 62.9 | 63.1 | 62.2 | 58.7 | 64.7 | 66.0 |
| 60.5 | 64.7 | 65.4 | 60.2 | 65.0 | 64.1 | 61.1 | 65.3 |
| 64.6 | 59.2 | 61.4 | 62.0 | 63.5 | 61.4 | 65.5 | 62.3 |
| 65.5 | 64.7 | 58.8 | 66.1 | 64.9 | 66.9 | 57.9 | 69.8 |
| 58.5 | 63.4 | 69.2 | 65.9 | 62.2 | 60.0 | 58.1 | 62.5 |
| 62.4 | 59.1 | 66.4 | 61.2 | 60.4 | 58.7 | 66.7 | 67.5 |
| 63.2 | 56.6 | 67.7 | 62.5 |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean (EBM). A confidence interval has the general form:
(lower bound, upper bound) = (point estimate – EBM, point estimate + EBM)
The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem,
Given a confidence interval, you can work backwards to find the error bound (EBM) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean.
Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal:
In many cases, the researcher does not know the population standard deviation, σ, of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of σ. The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
The t-score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = where is the t-score with area to the right equal to , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find for a given α.
Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning.
Let p′ represent the sample proportion, x/n, where x represents the number of successes and n represents the sample size. Let q′ = 1 – p′. Then the confidence interval for a population proportion is given by the following formula:
(lower bound, upper bound)
The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples.
The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.
The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
EBM = = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.
CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter
α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter
= the z-score with the property that the area to the right of the z-score is this is the z-score used in the calculation of "EBM where α = 1 – CL.
n = = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence
General form of a confidence interval
(lower value, upper value) = (point estimate−error bound, point estimate + error bound)
To find the error bound when you know the confidence interval
error bound = upper value−point estimate OR error bound =
Single Population Mean, Known Standard Deviation, Normal Distribution
Use the Normal Distribution for Means, Population Standard Deviation is Known EBM = z
The confidence interval has the format ( − EBM, + EBM).
s = the standard deviation of sample values.
is the formula for the t-score which measures how far away a measure is from the population mean in the Student’s t-distribution
df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom
= the error bound for the population mean when the population standard deviation is unknown
is the t-score in the Student’s t-distribution with area to the right equal to
The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound)
p′ = x / n where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion.
q′ = 1 – p′
The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here.
EBP = the error bound for a proportion =
Confidence interval for a proportion:
provides the number of participants needed to estimate the population proportion with confidence 1 - α and margin of error EBP.
Use the normal distribution for a single population proportion
The confidence interval has the format (p′ – EBP, p′ + EBP).
is a point estimate for μ
p′ is a point estimate for ρ
s is a point estimate for σ
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.
Identify the following:
In words, define the random variables X and .
Which distribution should you use for this problem?
Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound.
What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?
Identify the following:
In words, define the random variables X and .
Which distribution should you use for this problem?
Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound.
If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?
If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why?
Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not?
Identify the following:
In words, define the random variable X.
In words, define the random variable .
Which distribution should you use for this problem?
Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
In complete sentences, explain why the confidence interval in Exercise 8.17 is larger than in Exercise 8.18.
In complete sentences, give an interpretation of what the interval in Exercise 8.18 means.
What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same?
What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same?
= _____
n = _____
________ = 15
In words, define the random variable .
What is estimating?
Is known?
As a result of your answer to Exercise 8.26, state the exact distribution to use when calculating the confidence interval.
Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.
How much area is in both tails (combined)? α =________
How much area is in each tail? =________
Identify the following specifications:
The 95% confidence interval is:__________________.
Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean.
In one complete sentence, explain what the interval means.
Using the same mean, standard deviation, and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know?
Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?
Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.
Identify the following:
Define the random variables X and in words.
Which distribution should you use for this problem?
Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
Explain in complete sentences what the confidence interval means.
Identify the following:
Define the random variable X in words.
Define the random variable in words.
Which distribution should you use for this problem?
Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.
Why would the error bound change if the confidence level were lowered to 95%?
| X | Freq. |
|---|---|
| 1 | 1 |
| 2 | 7 |
| 3 | 18 |
| 4 | 7 |
| 5 | 6 |
Calculate the following:
Define the random variable in words.
What is estimating?
Is known?
As a result of your answer to Exercise 8.52, state the exact distribution to use when calculating the confidence interval.
How much area is in both tails (combined)?
How much area is in each tail?
Calculate the following:
The 95% confidence interval is_____.
Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean.
In one complete sentence, explain what the interval means.
Using the same , , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
Using the same , , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why?
Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions.
When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05?
If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why?
Identify the following:
Define the random variables X and P′ in words.
Which distribution should you use for this problem?
Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound.
List two difficulties the company might have in obtaining random results, if this survey were done by email.
We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P′ in words.
Which distribution should you use for this problem?
Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound.
Suppose we want to lower the sampling error. What is one way to accomplish that?
The sampling error given in the survey is ±2%. Explain what the ±2% means.
Define the random variable X in words.
Define the random variable P′ in words.
Which distribution should you use for this problem?
Construct a 90% confidence interval, and state the confidence interval and the error bound.
What would happen to the confidence interval if the level of confidence were 95%?
What is being counted?
In words, define the random variable X.
Calculate the following:
State the estimated distribution of X. X~________
Define a new random variable P′. What is p′ estimating?
In words, define the random variable P′.
State the estimated distribution of P′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet.
How much area is in both tails (combined)?
How much area is in each tail?
Calculate the following:
The 92% confidence interval is _______.
Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion.
In one complete sentence, explain what the interval means.
Using the same p′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know?
Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why?
If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)?
Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.
Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal.
Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.
A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.
A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.
What is meant by the term “90% confident” when constructing a confidence interval for a mean?
The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 8.11 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200.
| $3,600 | $1,243,900 | $10,900 | $385,200 | $581,500 |
| $7,400 | $2,900 | $400 | $3,714,500 | $632,500 |
| $391,000 | $467,400 | $56,800 | $5,800 | $405,200 |
| $733,200 | $8,000 | $468,700 | $75,200 | $41,000 |
| $13,300 | $9,500 | $953,800 | $1,113,500 | $1,109,300 |
| $353,900 | $986,100 | $88,600 | $378,200 | $13,200 |
| $3,800 | $745,100 | $5,800 | $3,072,100 | $1,626,700 |
| $512,900 | $2,309,200 | $6,600 | $202,400 | $15,800 |
The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The most recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income.
The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure?
In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces.
A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.
Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours.
A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.
Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.
The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.
The FEC has reported financial information for 556 Leadership PACs that operating during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs.
| $46,500.00 | $0 | $40,966.50 | $105,887.20 | $5,175.00 |
| $29,050.00 | $19,500.00 | $181,557.20 | $31,500.00 | $149,970.80 |
| $2,555,363.20 | $12,025.00 | $409,000.00 | $60,521.70 | $18,000.00 |
| $61,810.20 | $76,530.80 | $119,459.20 | $0 | $63,520.00 |
| $6,500.00 | $502,578.00 | $705,061.10 | $708,258.90 | $135,810.00 |
| $2,000.00 | $2,000.00 | $0 | $1,287,933.80 | $219,148.30 |
Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution.
Forbes magazine published data on the best small firms in 2012. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The Table 8.13 shows the ages of the corporate CEOs for a random sample of these firms.
| 48 | 58 | 51 | 61 | 56 |
| 59 | 74 | 63 | 53 | 50 |
| 59 | 60 | 60 | 57 | 46 |
| 55 | 63 | 57 | 47 | 55 |
| 57 | 43 | 61 | 62 | 49 |
| 67 | 67 | 55 | 55 | 49 |
Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution.
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.
In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal.
Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal.
A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.
Find the 95% Confidence Interval for the true population mean for the amount of soda served.
What is the error bound?
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.
Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job.
An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
Refer to the information in Exercise 8.120.
Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.
A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem.
Refer to Exercise 8.123. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
A point estimate for the true population proportion is:
A 90% confidence interval for the population proportion is _______.
The error bound is approximately _____.
Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.
The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.
On May 23, 2013, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error.
A national survey of 1,000 adults was conducted on May 13, 2013 by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education.
Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music.
You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview?
In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem.
“American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t (accessed July 2, 2013).
“Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013).
“Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm (accessed September 30,2013).
Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/2000growthchart-us.pdf (accessed July 2, 2013).
La, Lynn, Kent German. "Cell Phone Radiation Levels." c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013).
“Mean Income in the Past 12 Months (in 2011 Inflaction-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013).
“Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013).
“National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013).
“America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013).
Data from Microsoft Bookshelf.
Data from http://www.businessweek.com/.
Data from http://www.forbes.com/.
“Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013).
“Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013).
“Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013).
Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling. Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013).
Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, and Meredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx (accessed July 2, 2013).
Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.” Pew Research Center: Internet and American Life Project. Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013).
Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at http://www.gallup.com/poll/162758/three-four-workers-plan-work-past-retirement-age.aspx (accessed July 2, 2013).
The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2, 2013).
Zogby. “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. Available online at http://www.zogbyanalytics.com/news/299-americans-neither-worried-nor-prepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013).
“52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process (accessed July 2, 2013).
As the sample size increases, there will be less variability in the mean, so the interval size decreases.
X is the time in minutes it takes to complete the U.S. Census short form. is the mean time it took a sample of 200 people to complete the U.S. Census short form.
The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.
is the mean weight of a sample of 20 heads of lettuce.
The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.
The confidence level would increase.
30.4
σ
μ
normal
0.025
(24.52,36.28)
We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28.
The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.
X is the number of hours a patient waits in the emergency room before being called back to be examined. is the mean wait time of 70 patients in the emergency room.
is the mean number of hours spent watching television per month from a sample of 108 Americans.
μ
t38
0.025
(2.93, 3.59)
We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.
The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.
It would decrease, because the z-score would decrease, which reducing the numerator and lowering the number.
X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.
X is the number of “successes” where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck.
The sampling error means that the true mean can be 2% above or below the sample mean.
P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.
The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.
p
. (0.72171, 0.87829).
0.04
(0.72; 0.88)
With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%.
The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.
N
Alternate solution:
STAT and arrow over to TESTS.7:ZInterval.ENTER.ENTER.ENTER.Use the formula for EBM, solved for n:
From the statement of the problem, you know that σ = 2.5, and you need EBM = 1.
z = z0.035 = 1.812
(This is the value of z for which the area under the density curve to the right of z is 0.035.)
You need to measure at least 21 male students to achieve your goal.
Note that we are not given the population standard deviation, only the standard deviation of the sample.
There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29
CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04
= 2.150
- EBM = $251,854.23 - $204,561.66 = $47,292.57
+ EBM = $251,854.23+ $204,561.66 = $456,415.89
We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.
Alternate Solution
Enter the data as a list.
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval.
Press ENTER.
Arrow to Data and press ENTER.
Arrow down and enter the name of the list where the data is stored.
Enter Freq: 1
Enter C-Level: 0.96
Arrow down to Calculate and press Enter.
The 96% confidence interval is ($47,262, $456,447).
The difference between solutions arises from rounding differences.
b
X = the number of people who feel that the president is doing an acceptable job;
P′ = the proportion of people in a sample who feel that the president is doing an acceptable job.
c
d
a
Alternate Solution
STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.
Answer is (0.502, 0.538)
CL = 0.95 α = 1 – 0.95 = 0.05 = 0.025 = 1.96. Use p′ = q′ = 0.5.
You need to interview at least 385 students to estimate the proportion to within 5% at 95% confidence.
By the end of this chapter, the student should be able to:
One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.
A statistician will make a decision about these claims. This process is called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis.
In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.
Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will:
To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Solution Sheets.
The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints.
H0: The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.
Ha: The alternative hypothesis: It is a claim about the population that is contradictory to H0 and what we conclude when we reject H0.
Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.
After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject H0" if the sample information favors the alternative hypothesis or "do not reject H0" or "decline to reject H0" if the sample information is insufficient to reject the null hypothesis.
Mathematical Symbols Used in H0 and Ha:
| H0 | Ha |
|---|---|
| equal (=) | not equal (≠) or greater than (>) or less than (<) |
| greater than or equal to (≥) | less than (<) |
| less than or equal to (≤) | more than (>) |
H0 always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.
H0: No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30
A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.
H0 : The drug reduces cholesterol by 25%. p = 0.25
Ha : The drug does not reduce cholesterol by 25%. p ≠ 0.25
We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:
We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:
We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
In an issue of U. S. News and World Report, an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.
On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.
When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis H0 and the decision to reject or not. The outcomes are summarized in the following table:
| ACTION | H0 IS ACTUALLY | ... |
|---|---|---|
| True | False | |
| Do not reject H0 | Correct Outcome | Type II error |
| Reject H0 | Type I Error | Correct Outcome |
The four possible outcomes in the table are:
Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.
α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true.
β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false.
α and β should be as small as possible because they are probabilities of errors. They are rarely zero.
The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test.
The following are examples of Type I and Type II errors.
Suppose the null hypothesis, H0, is: Frank's rock climbing equipment is safe.
Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe. Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.
α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe.
Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)
Suppose the null hypothesis, H0, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors.
Type I error: The researcher thinks the blood cultures do contain traces of pathogen X, when in fact, they do not.
Type II error: The researcher thinks the blood cultures do not contain traces of pathogen X, when in fact, they do.
Suppose the null hypothesis, H0, is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital.
Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead.
α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P(Type I error). β = probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P(Type II error).
The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)
Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?
The error with the greater consequence is the Type II error: the patient will be thought well when, in fact, he is sick, so he will not get treatment.
It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, H0, is: It’s a Boy Genetic Labs has no effect on gender outcome.
Type I error: This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α.
Type II error: This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β.
The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy.
“Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.
In this scenario, an appropriate null hypothesis would beH0: the mean level of toxins is at most 800 μg, H0 : μ0 ≤ 800 μg.
Type I error: The DMF believes that toxin levels are still too high when, in fact, toxin levels are at most 800 μg. The DMF continues the harvesting ban.
Type II error: The DMF believes that toxin levels are within acceptable levels (are at least 800 μg) when, in fact, toxin levels are still too high (more than 800 μg). The DMF lifts the harvesting ban. This error could be the most serious. If the ban is lifted and clams are still toxic, consumers could possibly eat tainted food.
In summary, the more dangerous error would be to commit a Type II error, because this error involves the availability of tainted clams for consumption.
A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious?
Type I: A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%.
Type II: A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%.
In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option.
Determine both Type I and Type II errors for the following scenario:
Assume a null hypothesis, H0, that states the percentage of adults with jobs is at least 88%.
Identify the Type I and Type II errors from these four statements.
Type I error: c
Type I error: b
Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large or the sample size is large).
If you are testing a single population mean, the distribution for the test is for means:
or
The population parameter is μ. The estimated value (point estimate) for μ is , the sample mean.
If you are testing a single population proportion, the distribution for the test is for proportions or percentages:
The population parameter is p. The estimated value (point estimate) for p is p′. p′ = where x is the number of successes and n is the sample size.
When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed).
When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.
When you perform a hypothesis test of a single population proportion p, you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and . Remember that q = 1 – p.
Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.
Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption—it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.)
For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket.
Use the sample data to calculate the actual probability of getting the test result, called the p-value. The p-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample.
A large p-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the p-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it.
Draw a graph that shows the p-value. The hypothesis test is easier to perform if you use a graph because you see the problem more clearly.
Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal.
The null hypothesis could be H0: μ ≤ 15 The alternate hypothesis is Ha: μ > 15
The words "is more than" translates as a ">" so "μ > 15" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis.
Since σ is known (σ = 0.5 cm.), the distribution for the population is known to be normal with mean μ = 15 and standard deviation .
Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm.
The p-value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means.
p-value= P( > 17) which is approximately zero.
A p-value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone), we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population of the baker's loaves of bread is greater than 15 cm.
A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5.
H0: μ ≤ 12
p-value = 0.0013
A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p-value and a preset or preconceived α (also called a "significance level"). A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.
When you make a decision to reject or not reject H0, do as follows:
Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.
When using the p-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device
If the p-value is low, the null must go.
If the p-value is high, the null must fly.
This memory aid relates a p-value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p-value higher than the established alpha (the p is high) as not rejecting the null hypothesis.
Fill in the blanks.
Reject the null hypothesis when ______________________________________.
The results of the sample data _____________________________________.
Do not reject the null when hypothesis when __________________________________________.
The results of the sample data ____________________________________________.
Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis.
Do not reject the null hypothesis when the p-value is greater than the established alpha value. The results of the sample data do not support the alternative hypothesis.
It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows:
H0: p = 0.50, Ha: p > 0.50
α = 0.01
p-value = 0.025
Interpret the results and state a conclusion in simple, non-technical terms.
Since the p-value is greater than the established alpha value (the p-value is high), we do not reject the null hypothesis. There is not enough evidence to support It’s a Boy Genetics Labs' stated claim that their procedures improve the chances of a boy being born.
The following examples illustrate a left-, right-, and two-tailed test.
Ho: μ = 5, Ha: μ < 5
Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows:
H0: μ = 10, Ha: μ < 10
Assume the p-value is 0.0935. What type of test is this? Draw the picture of the p-value.
left-tailed test
H0: p ≤ 0.2 Ha: p > 0.2
This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of the p-value is as follows:
H0: μ ≤ 1, Ha: μ > 1
Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value.
right-tailed test
H0: p = 50 Ha: p ≠ 50
This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the p-value is as follows.
H0: p = 0.5, Ha: p ≠ 0.5
Assume the p-value is 0.2564. What type of test is this? Draw the picture of the p-value.
two-tailed test
Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.
Set up the Hypothesis Test:
Since the problem is about a mean, this is a test of a single population mean.
H0: μ = 16.43 Ha: μ < 16.43
For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.
Determine the distribution needed:
Random variable: = the mean time to swim the 25-yard freestyle.
Distribution for the test: is normal (population standard deviation is known: σ = 0.8)
Therefore,
μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15.
Calculate the p-value using the normal distribution for a mean:
p-value = P( < 16) = 0.0187 where the sample mean in the problem is given as 16.
p-value = 0.0187 (This is called the actual level of significance.) The p-value is the area to the left of the sample mean is given as 16.
Graph:
μ = 16.43 comes from H0. Our assumption is μ = 16.43.
Interpretation of the p-value: If H0 is true, there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.
Compare α and the p-value:
α = 0.05 p-value = 0.0187 α > p-value
Make a decision: Since α > p-value, reject H0.
This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.
Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.
The p-value can easily be calculated.
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw(instead of Calculate). Press ENTER. A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
When the calculator does a Z-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the central limit theorem:
2nd DISTR normcdf
.
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n. Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select <, ≠, or > for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and p-value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
Since the problem is about a mean, this is a test of a single population mean.
H0 : μ = 40
Ha : μ > 40
p = 0.0062
Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the Central Limit Theorem, the test statistic formula is . For this problem, = 16, μX = 16.43 from the null hypothes is, σX = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see 15.Tables in the Table of Contents). The z-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very common. For this problem, the p-value, 0.0187 is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p-value, and the test statistics and the critical value.
A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3) 215(3) 225(1) 241(2) 252(2) 265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1).
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Set up the Hypothesis Test:
Since the problem is about a mean weight, this is a test of a single population mean.
H0: μ = 275
Calculating the distribution needed:
Random variable: = the mean weight, in pounds, lifted by the football players.
Distribution for the test: It is normal because σ is known.
pounds (from the data).
σ = 55 pounds (Always use σ if you know it.) We assume μ = 275 pounds unless our data shows us otherwise.
Calculate the p-value using the normal distribution for a mean and using the sample mean as input (see Table 13.45 for using the data as input):
.
Interpretation of the p-value: If H0 is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
Compare α and the p-value:
α = 0.025 p-value = 0.1323
Make a decision: Since α <p-value, do not reject H0.
Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
The p-value can easily be calculated.
Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0, 55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.1331, a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.
Set up the hypothesis test:
A 5% level of significance means that α = 0.05. This is a test of a single population mean.
H0: μ = 65 Ha: μ > 65
Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed.
Determine the distribution needed:
Random variable: = average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's t.
Use tdf. Therefore, the distribution for the test is t9 where n = 10 and df = 10 - 1 = 9.
Calculate the p-value using the Student's t-distribution:
p-value = P( > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.
Compare α and the p-value:
Since α = 0.05 and p-value = 0.0396. α > p-value.
Make a decision: Since α > p-value, reject H0.
This means you reject μ = 65. In other words, you believe the average test score is more than 65.
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The p-value can easily be calculated.
Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.
H0: μ = 5
Ha: μ < 5
p = 0.0082
Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.
Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).
Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.
Set up the hypothesis test:
The 1% level of significance means that α = 0.01. This is a test of a single population proportion.
H0: p = 0.50 Ha: p ≠ 0.50
The words "is the same or different from" tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: P′ = the percent of of first-time brides who are younger than their grooms.
Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion.
Therefore,
where p = 0.50, q = 1−p = 0.50, and n = 100
Calculate the p-value using the normal distribution for proportions:
p-value = P (p′ < 0.47 or p′ > 0.53) = 0.5485
where x = 53, p′ = = 0.53.
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion is 0.53 or more OR 0.47 or less (see the graph in Figure 9.15).
μ = p = 0.50 comes from H0, the null hypothesis.
p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.)
Compare α and the p-value:
Since α = 0.01 and p-value = 0.5485. α < p-value.
Make a decision: Since α < p-value, you cannot reject H0.
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.
The p-value can easily be calculated.
Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for p0, 53 for x and 100 for n. Arrow down to Prop and arrow to not equals p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p = 0.5485) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).
The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
Since the problem is about percentages, this is a test of single population proportions.
H0 : p = 0.85
Ha: p ≠ 0.85
p = 0.7554
Because p > α, we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.
Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.
Set up the Hypothesis Test:
H0: p = 0.30 Ha: p ≠ 0.30
Determine the distribution needed:
The random variable is P′ = proportion of households that have three cell phones.
The distribution for the hypothesis test is
a. The value that helps determine the p-value is p′. Calculate p′.
a. p′ = where x is the number of successes and n is the total number in the sample.
x = 43, n = 150
p′ =
b. What is a success for this problem?
c. What is the level of significance?
c. The level of significance is the preset α. Since α is not given, assume that α = 0.05.
d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
e. Make a decision. _____________(Reject/Do not reject) H0 because____________.
e. Assuming that α = 0.05, α < p-value. The decision is do not reject H0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.
Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p-value, state your conclusion, and identify the Type I and Type II errors.
H0: p = 0.92
Ha: p < 0.92
p-value = 0.0046
Because p < 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.
Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).
The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!
My dog has so many fleas,
Set up the hypothesis test:
H0: p ≤ 0.25 Ha: p > 0.25
Determine the distribution needed:
In words, CLEARLY state what your random variable or P′ represents.
P′ = The proportion of fleas that are killed by the new shampoo
State the distribution to use for the test.
Normal:
Test Statistic: z = 2.3163
Calculate the p-value using the normal distribution for proportions:
p-value = 0.0103
In one to two complete sentences, explain what the p-value means for this problem.
If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 or more.
Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.
Compare α and the p-value:
Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.
| alpha | decision | reason for decision |
|---|---|---|
| 0.01 | Do not reject | α < p-value |
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.
Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.
Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.
This test result is not very definitive since the p-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.
The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.
1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95
Let’s follow a four-step process to answer this statistical question.
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
We will follow the four-step process.
If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.
According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.
We will follow the four-step plan.
Class Time:
Names:
Television SurveyIn a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower.
Language SurveyAbout 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%.
Jeans SurveySuppose that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal.
In a hypothesis test, sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we:
In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected.
The probabilities of these errors are denoted by the Greek letters α and β, for a Type I and a Type II error respectively. The power of the test, 1 – β, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable.
In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.
When testing for a single population mean:
When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure.
When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:
The hypothesis test itself has an established process. This can be summarized as follows:
Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that is used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.
H0 and Ha are contradictory.
| If Ho has: | equal (=) | greater than or equal to (≥) | less than or equal to (≤) |
| then Ha has: | not equal (≠) or greater than (>) or less than (<) | less than (<) | greater than (>) |
If α ≤ p-value, then do not reject H0.
If α > p-value, then reject H0.
α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data.
α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true.
β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false.
If there is no given preconceived α, then use α = 0.05.
You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words.
You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses.
The American family has an average of two children. What is the random variable? Describe in words.
The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses.
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words.
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses.
In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses.
Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal.
A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time on death row could likely be 15 years, what would the null and alternative hypotheses be?
The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population, what would the null and alternative hypotheses be?
The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences.
A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences.
For Exercise 9.12, what are α and β in words?
In words, describe 1 – β For Exercise 9.12.
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences.
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. Which is the error with the greater consequence?
The power of a test is 0.981. What is the probability of a Type II error?
A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H0, is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences.
A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H0, is: the sample does not contain E-coli. The probability that the sample does not contain E-coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E-coli, but the microbiologist thinks it does not is 0.002. What is the power of this test?
A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H0, is: the sample contains E-coli. Which is the error with the greater consequence?
Which two distributions can you use for hypothesis testing for this chapter?
Which distribution do you use when you are testing a population mean and the standard deviation is known? Assume sample size is large.
Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large.
A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal.
A population has a mean is 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test?
It is thought that 42% of respondents in a taste test would prefer Brand A. In a particular test of 100 people, 39% preferred Brand A. What distribution should you use to perform a hypothesis test?
You are performing a hypothesis test of a single population mean using a Student’s t-distribution. What must you assume about the distribution of the data?
You are performing a hypothesis test of a single population mean using a Student’s t-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test?
You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq?
You are performing a hypothesis test of a single population proportion. You find out that np is less than five. What must you do to be able to perform a valid hypothesis test?
You are performing a hypothesis test of a single population proportion. The data come from which distribution?
When do you reject the null hypothesis?
The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely?
The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why?
It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 years. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value.
The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The p-value is 0.0264. State the null and alternative hypotheses and interpret the p-value.
Does the shaded region represent a low or a high p-value compared to a level of significance of 1%?
What should you do when α > p-value?
What should you do if α = p-value?
If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences.
Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Is this a test of means or proportions?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
Is the population standard deviation known and, if so, what is it?
Calculate the following:
Since both σ and are given, which should be used? In one to two complete sentences, explain why.
State the distribution to use for the hypothesis test.
A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years.
Assume H0: μ = 9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test?
Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test?
Assume H0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test?
Draw the general graph of a left-tailed test.
Draw the graph of a two-tailed test.
A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?
Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?
A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?
You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?
If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test?
Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?
Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?
Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?
Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, H0, and the alternative hypothesis. Ha, in terms of the appropriate parameter (μ or p).
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:
State the Type I and Type II errors in complete sentences given the following statements.
For statements a-j in Exercise 9.109, answer the following in complete sentences.
When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?
The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is ~ ________________
The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.
For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Table 13.45. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
If you are using a Student's-t distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?
From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?
In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?
Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?
A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.
Refer to Exercise 9.119. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.
According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?
A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.
The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?
Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
Use the “Lap time” data for Lap 4 (see Table 13.45) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.
Use the “Initial Public Offering” data (see Table 13.45) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.
The following questions were written by past students. They are excellent problems!
"Asian Family Reunion," by Chau Nguyen
Every two years it comes around.
We all get together from different towns.
In my honest opinion,
It's not a typical family reunion.
Not forty, or fifty, or sixty,
But how about seventy companions!
The kids would play, scream, and shout
One minute they're happy, another they'll pout.
The teenagers would look, stare, and compare
From how they look to what they wear.
The men would chat about their business
That they make more, but never less.
Money is always their subject
And there's always talk of more new projects.
The women get tired from all of the chats
They head to the kitchen to set out the mats.
Some would sit and some would stand
Eating and talking with plates in their hands.
Then come the games and the songs
And suddenly, everyone gets along!
With all that laughter, it's sad to say
That it always ends in the same old way.
They hug and kiss and say "good-bye"
And then they all begin to cry!
I say that 60 percent shed their tears
But my mom counted 35 people this year.
She said that boys and men will always have their pride,
So we won't ever see them cry.
I myself don't think she's correct,
So could you please try this problem to see if you object?
"The Problem with Angels," by Cyndy Dowling
Although this problem is wholly mine,
The catalyst came from the magazine, Time.
On the magazine cover I did find
The realm of angels tickling my mind.
Inside, 69% I found to be
In angels, Americans do believe.
Then, it was time to rise to the task,
Ninety-five high school and college students I did ask.
Viewing all as one group,
Random sampling to get the scoop.
So, I asked each to be true,
"Do you believe in angels?" Tell me, do!
Hypothesizing at the start,
Totally believing in my heart
That the proportion who said yes
Would be equal on this test.
Lo and behold, seventy-three did arrive,
Out of the sample of ninety-five.
Now your job has just begun,
Solve this problem and have some fun.
"Blowing Bubbles," by Sondra Prull
Studying stats just made me tense,
I had to find some sane defense.
Some light and lifting simple play
To float my math anxiety away.
Blowing bubbles lifts me high
Takes my troubles to the sky.
POIK! They're gone, with all my stress
Bubble therapy is the best.
The label said each time I blew
The average number of bubbles would be at least 22.
I blew and blew and this I found
From 64 blows, they all are round!
But the number of bubbles in 64 blows
Varied widely, this I know.
20 per blow became the mean
They deviated by 6, and not 16.
From counting bubbles, I sure did relax
But now I give to you your task.
Was 22 a reasonable guess?
Find the answer and pass this test!
"Dalmatian Darnation," by Kathy Sparling
A greedy dog breeder named Spreckles
Bred puppies with numerous freckles
The Dalmatians he sought
Possessed spot upon spot
The more spots, he thought, the more shekels.
His competitors did not agree
That freckles would increase the fee.
They said, “Spots are quite nice
But they don't affect price;
One should breed for improved pedigree.”
The breeders decided to prove
This strategy was a wrong move.
Breeding only for spots
Would wreak havoc, they thought.
His theory they want to disprove.
They proposed a contest to Spreckles
Comparing dog prices to freckles.
In records they looked up
One hundred one pups:
Dalmatians that fetched the most shekels.
They asked Mr. Spreckles to name
An average spot count he'd claim
To bring in big bucks.
Said Spreckles, “Well, shucks,
It's for one hundred one that I aim.”
Said an amateur statistician
Who wanted to help with this mission.
“Twenty-one for the sample
Standard deviation's ample:
They examined one hundred and one
Dalmatians that fetched a good sum.
They counted each spot,
Mark, freckle and dot
And tallied up every one.
Instead of one hundred one spots
They averaged ninety six dots
Can they muzzle Spreckles’
Obsession with freckles
Based on all the dog data they've got?
"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall
As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.
One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:
I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)
"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi
Scene: The great library of the castle, in which Hamlet does his lessons
Act I
(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)
POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!
HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.
POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.
HORATIO (to Hamlet): What should we do, my Lord?
HAMLET: Go to thy purpose, Horatio.
HORATIO: To what end, my Lord?
HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.
(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)
Act II
(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)
POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?
HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.
POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)
HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)
(Curtain falls)
"Untitled," by Stephen Chen
I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.
So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?
"Japanese Girls’ Names"
by Kumi Furuichi
It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.
However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.
I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.
Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.
"Phillip’s Wish," by Suzanne Osorio
My nephew likes to play
Chasing the girls makes his day.
He asked his mother
If it is okay
To get his ear pierced.
She said, “No way!”
To poke a hole through your ear,
Is not what I want for you, dear.
He argued his point quite well,
Says even my macho pal, Mel,
Has gotten this done.
It’s all just for fun.
C’mon please, mom, please, what the hell.
Again Phillip complained to his mother,
Saying half his friends (including their brothers)
Are piercing their ears
And they have no fears
He wants to be like the others.
She said, “I think it’s much less.
We must do a hypothesis test.
And if you are right,
I won’t put up a fight.
But, if not, then my case will rest.”
We proceeded to call fifty guys
To see whose prediction would fly.
Nineteen of the fifty
Said piercing was nifty
And earrings they’d occasionally buy.
Then there’s the other thirty-one,
Who said they’d never have this done.
So now this poem’s finished.
Will his hopes be diminished,
Or will my nephew have his fun?
"The Craven," by Mark Salangsang
Once upon a morning dreary
In stats class I was weak and weary.
Pondering over last night’s homework
Whose answers were now on the board
This I did and nothing more.
While I nodded nearly napping
Suddenly, there came a tapping.
As someone gently rapping,
Rapping my head as I snore.
Quoth the teacher, “Sleep no more.”
“In every class you fall asleep,”
The teacher said, his voice was deep.
“So a tally I’ve begun to keep
Of every class you nap and snore.
The percentage being forty-four.”
“My dear teacher I must confess,
While sleeping is what I do best.
The percentage, I think, must be less,
A percentage less than forty-four.”
This I said and nothing more.
“We’ll see,” he said and walked away,
And fifty classes from that day
He counted till the month of May
The classes in which I napped and snored.
The number he found was twenty-four.
At a significance level of 0.05,
Please tell me am I still alive?
Or did my grade just take a dive
Plunging down beneath the floor?
Upon thee I hereby implore.
Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.
Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.
According to an article in Bloomberg Businessweek, New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.
The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.
Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.
La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
Instructions: For the following ten exercises,
According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?
Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate?
The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level in Kentucky? Are the results applicable across the country? Why?
For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?
The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution.
A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?
A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals
3 2 1 3 7 2 9 4 6 6 8 0 5 6 4 2 1 3 4 1
According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:
The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?
Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm.
Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
Data from Bloomberg Businessweek. Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
Data from Growing by Degrees by Allen and Seaman.
Data from La Leche League International. Available online at http://www.lalecheleague.org/Law/BAFeb01.html.
Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm.
Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
Data from the U.S. Census Bureau, available online at http://quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
Data from the United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/.
Data from Toastmasters International. Available online at http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1.
Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
“Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf.
Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at http://www.rainn.org/get-information/statistics/frequency-of-sexual-assault (accessed June 27, 2013).
The random variable is the mean Internet speed in Megabits per second.
The random variable is the mean number of children an American family has.
The random variable is the proportion of people picked at random in Times Square visiting the city.
Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000.
Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000.
α = the probability that you think the bag cannot withstand -15 degrees F, when in fact it can
β = the probability that you think the bag can withstand -15 degrees F, when in fact it cannot
Type I: The procedure will go well, but the doctors think it will not.
Type II: The procedure will not go well, but the doctors think it will.
0.019
0.998
A normal distribution or a Student’s t-distribution
Use a Student’s t-distribution
a normal distribution for a single population mean
It must be approximately normally distributed.
They must both be greater than five.
The outcome of winning is very unlikely.
H0: μ > = 73
The shaded region shows a low p-value.
Do not reject H0.
means
the mean time spent in jail for 26 first time convicted burglars
This is a left-tailed test.
This is a two-tailed test.
a right-tailed test
a left-tailed test
This is a left-tailed test.
This is a two-tailed test.
c
b
d
d
Note that here the “large-sample” 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close to the level of significance, the exact p-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.
c
c
By the end of this chapter, the student should be able to:
Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years.
There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores.
You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded.
To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.
This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p-values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests.
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, – , and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.
Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, – .
The test statistic (t-score) is calculated as follows:
The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student's t-distribution with df as follows:
When both sample sizes n1 and n2 are five or larger, the Student's t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.)
The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 10.1. Each populations has a normal distribution.
| Sample Size | Average Number of Hours Playing Sports Per Day | Sample Standard Deviation | |
|---|---|---|---|
| Girls | 9 | 2 | |
| Boys | 16 | 3.2 | 1.00 |
Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.
The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means.
Random variable: = difference in the sample mean amount of time girls and boys play sports each day.
Distribution for the test: Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.
Calculate the p-value using a Student's t-distribution: p-value = 0.0054
Graph:
Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different.
Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean,
for Sx1, 9 for n1, 3.2 for the
second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.
Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).
Two samples are shown in Table 10.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.
| Sample Size | Sample Mean | Sample Standard Deviation | |
|---|---|---|---|
| Population A | 25 | 5 | 1 |
| Population B | 16 | 4.7 | 1.2 |
The p-value is 0.4125, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.
When the sum of the sample sizes is larger than 30 (n1 + n2 > 30) you can use the normal distribution to approximate the Student's t.
A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.
a. Is this a test of two means or two proportions?
b. Are the populations standard deviations known or unknown?
c. Which distribution do you use to perform the test?
d. What is the random variable?
e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
f. Is this test right-, left-, or two-tailed?
g. What is the p-value?
h. Do you reject or not reject the null hypothesis?
i. Conclusion:
i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.
A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.
A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 10.3 and Table 10.4.
| 67.6 | 41.2 | 85.3 | 55.9 | 82.4 | 91.2 | 73.5 | 94.1 | 64.7 | 64.7 |
| 70.6 | 38.2 | 61.8 | 88.2 | 70.6 | 58.8 | 91.2 | 73.5 | 82.4 | 35.5 |
| 94.1 | 88.2 | 64.7 | 55.9 | 88.2 | 97.1 | 85.3 | 61.8 | 79.4 | 79.4 |
| 77.9 | 95.3 | 81.2 | 74.1 | 98.8 | 88.2 | 85.9 | 92.9 | 87.1 | 88.2 |
| 69.4 | 57.6 | 69.4 | 67.1 | 97.6 | 85.9 | 88.2 | 91.8 | 78.8 | 71.8 |
| 98.8 | 61.2 | 92.9 | 90.6 | 97.6 | 100 | 95.3 | 83.5 | 92.9 | 89.4 |
Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:
(See the conclusion in Example 10.1, and write yours in a similar fashion)
First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.
Be careful not to mix up the information for Group 1 and Group 2!
Cohen's Standards for Small, Medium, and Large Effect SizesCohen's d is a measure of effect size based on the differences between two means. Cohen’s d, named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.
| Size of effect | d |
|---|---|
| Small | 0.2 |
| medium | 0.5 |
| Large | 0.8 |
Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: where
Calculate Cohen’s d for Example 10.1. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.
μ1 = 4 s1 = 1.5 n1 = 11
Calculate Cohen’s d for Example 10.2. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.
d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.
Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table 10.6 and Table 10.7, respectively.
| 94.2 | 75.2 | 69.6 | 52.0 | 48.0 | 41.9 | 36.4 | 33.4 | 31.5 | 27.6 |
| 77.3 | 71.9 | 67.5 | 50.6 | 46.2 | 38.4 | 35.2 | 33.0 | 28.7 | 26.5 |
| 76.3 | 71.7 | 56.3 | 48.7 | 43.2 | 37.6 | 33.7 | 31.8 | 28.5 | 26.0 |
| 126.0 | 70.6 | 65.2 | 51.4 | 45.5 | 37.0 | 33.0 | 29.6 | 23.7 | 22.6 |
| 116.1 | 70.6 | 58.2 | 51.2 | 43.2 | 36.0 | 31.4 | 28.7 | 23.5 | 21.6 |
| 78.2 | 68.2 | 55.6 | 50.3 | 39.0 | 34.1 | 31.0 | 25.3 | 23.4 | 21.5 |
Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:
Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is . The normal distribution has the following format:
Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table 10.8.
| Wax | Sample Mean Number of Months Floor Wax Lasts | Population Standard Deviation |
|---|---|---|
| 1 | 3 | 0.33 |
| 2 | 2.9 | 0.36 |
Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.
This is a test of two independent groups, two population means, population standard deviations known.
Random Variable: = difference in the mean number of months the competing floor waxes last.
H0: μ1 ≤ μ2
Ha: μ1 > μ2
The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into Ha. Therefore, this is a right-tailed test.
Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is:
Since μ1 ≤ μ2 then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero.
Calculate the p-value using the normal distribution: p-value = 0.1799
Graph:
– = 3 – 2.9 = 0.1
Compare α and the p-value: α = 0.05 and p-value = 0.1799. Therefore, α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.
Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw.
The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.
| Engine | Sample Mean Number of RPM | Population Standard Deviation |
|---|---|---|
| 1 | 1,500 | 50 |
| 2 | 1,600 | 60 |
The p-value is almost zero, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1.
An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years.
Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance.
This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators
Random variable: = difference in the mean age of Democratic and Republican U.S. senators.
H0: µ1 ≤ µ2 H0: µ1 – µ2 ≤ 0
Ha: µ1 > µ2 Ha: µ1 – µ2 > 0
The words "older than" translates as a “>” symbol and goes into Ha. Therefore, this is a right-tailed test.
Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is:
Since µ1 ≤ µ2, µ1 – µ2 ≤ 0 and the mean for the normal distribution is zero.
(Calculating the p-value using the normal distribution gives p-value = 0.4040)
Graph:
Compare α and the p-value: α = 0.05 and p-value = 0.4040. Therefore, α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.
When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc.
Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.
The problem asks for a difference in proportions, making it a test of two proportions.
Let A and B be the subscripts for medication A and medication B, respectively. Then pA and pB are the desired population proportions.
Random Variable: P′A – P′B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.
H0: pA = pB
pA – pB = 0
Ha: pA ≠ pB
pA – pB ≠ 0
The words "is a difference" tell you the test is two-tailed.
Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:
P′A – P′B follows an approximate normal distribution.
Calculate the p-value using the normal distribution: p-value = 0.1404.
Estimated proportion for group A:
Estimated proportion for group B:
Graph:
P′A – P′B = 0.1 – 0.06 = 0.04.
Half the p-value is below –0.04, and half is above 0.04.
Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.
Press STAT. Arrow over to TESTS and press
6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2,
and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw.
Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance.
The p-value is 0.0379, so we can reject the null hypothesis. At the 5% significance level, the data support that there is a difference in the pressure tolerances between the two valves.
A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.10. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance.
| Males | Females | |
|---|---|---|
| Sent “sexts” | 183 | 156 |
| Total number surveyed | 2231 | 2169 |
This is a test of two population proportions. Let M and F be the subscripts for males and females. Then pM and pF are the desired population proportions.
Random variable: p′F − p′M = difference in the proportions of males and females who sent “sexts.”
H0: pF = pM H0: pF – pM = 0
Ha: pF < pM Ha: pF – pM < 0
The words "less than" tell you the test is left-tailed.
Distribution for the test: Since this is a test of two population proportions, the distribution is normal:
Calculate the p-value using the normal distribution:
Graph:
Decision: Since α < p-value, Do not reject H0
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.”
Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is P = 0.1045 and the test statistic is z = -1.256.
Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners?
This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then pW and pA are the desired population proportions.
Random variable:p′W – p′A = difference in the proportions of Android and iPhone users.
H0: pW = pA H0: pW – pA = 0
Ha: pW > pA Ha: pW – pA > 0
The words "more popular" indicate that the test is right-tailed.
Distribution for the test: The distribution is approximately normal:
Therefore,
follows an approximate normal distribution.
Calculate the p-value using the normal distribution:
Graph:
Decision: Since α > p-value, reject the H0.
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans.
TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The P-value is P = 0.0092 and the test statistic is Z = 2.33.
A concerned group of citizens wanted to know if the proportion of forcible rapes in Texas was different in 2011 than in 2010. Their research showed that of the 113,231 violent crimes in Texas in 2010, 7,622 of them were forcible rapes. In 2011, 7,439 of the 104,873 violent crimes were in the forcible rape category. Test at a 5% significance level. Answer the following questions:
a. Is this a test of two means or two proportions?
a. two proportions
b. Which distribution do you use to perform the test?
b. normal for two proportions
c. What is the random variable?
c. Subscripts: 1 = 2010, 2 = 2011
d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols.
d. Subscripts: 1 = 2010, 2 = 2011
e. Is this test right-, left-, or two-tailed?
e. two-tailed
f. What is the p-value?
f. p-value = 0.00086
g. Do you reject or not reject the null hypothesis?
g. Reject the H0.
h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________.
h. At the 5% significance level, from the sample data, there is sufficient evidence to conclude that there is a difference between the proportion of forcible rapes in 2011 and 2010.
When using a hypothesis test for matched or paired samples, the following characteristics should be present:
In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences.
A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.11. A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.
| Subject: | A | B | C | D | E | F | G | H |
|---|---|---|---|---|---|---|---|---|
| Before | 6.6 | 6.5 | 9.0 | 10.3 | 11.3 | 8.1 | 6.3 | 11.6 |
| After | 6.8 | 2.4 | 7.4 | 8.5 | 8.1 | 6.1 | 3.4 | 2.0 |
Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.")
| After Data | Before Data | Difference |
|---|---|---|
| 6.8 | 6.6 | 0.2 |
| 2.4 | 6.5 | -4.1 |
| 7.4 | 9 | -1.6 |
| 8.5 | 10.3 | -1.8 |
| 8.1 | 11.3 | -3.2 |
| 6.1 | 8.1 | -2 |
| 3.4 | 6.3 | -2.9 |
| 2 | 11.6 | -9.6 |
The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6}
The sample mean and sample standard deviation of the differences are: and Verify these values.
Let be the population mean for the differences. We use the subscript to denote "differences."
Random variable: = the mean difference of the sensory measurements
H0: μd ≥ 0
The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μd is the population mean of the differences.)
Ha: μd < 0
The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.
Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t7. (Notice that the test is for a single population mean.)
Calculate the p-value using the Student's-t distribution: p-value = 0.0095
Graph:
is the random variable for the differences.
The sample mean and sample standard deviation of the differences are:
= –3.13
= 2.91
Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value.
Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement.
Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.
For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after - before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list.
Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to < . Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER.
A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level.
| Subject | A | B | C | D | E | F | G | H | I |
| Before | 209 | 210 | 205 | 198 | 216 | 217 | 238 | 240 | 222 |
| After | 199 | 207 | 189 | 209 | 217 | 202 | 211 | 223 | 201 |
The p-value is 0.0130, so we can reject the null hypothesis. There is enough evidence to suggest that the diet lowers cholesterol.
A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:
| Weight (in pounds) | Player 1 | Player 2 | Player 3 | Player 4 |
|---|---|---|---|---|
| Amount of weight lifted prior to the class | 205 | 241 | 338 | 368 |
| Amount of weight lifted after the class | 295 | 252 | 330 | 360 |
The coach wants to know if the strength development class makes his players stronger, on average.
Using the differences data, calculate the sample mean and the sample standard deviation.
= 21.3, sd = 46.7
The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative.
Using the difference data, this becomes a test of a single __________ (fill in the blank).
Define the random variable: mean difference in the maximum lift per player.
The distribution for the hypothesis test is t3.
H0: μd ≤ 0, Ha: μd > 0
Graph:
Calculate the p-value: The p-value is 0.2150
Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p-value.
What is the conclusion?
At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.
A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5% level.
| SAT Scores | Student 1 | Student 2 | Student 3 | Student 4 |
|---|---|---|---|---|
| Score before class | 1840 | 1960 | 1920 | 2150 |
| Score after class | 1920 | 2160 | 2200 | 2100 |
The p-value is 0.0874, so we decline to reject the null hypothesis. The data do not support that the class improves SAT scores significantly.
Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in Table 10.16.
| Distance (in feet) using | Student 1 | Student 2 | Student 3 | Student 4 | Student 5 | Student 6 | Student 7 |
|---|---|---|---|---|---|---|---|
| Dominant Hand | 30 | 26 | 34 | 17 | 19 | 26 | 20 |
| Weaker Hand | 28 | 14 | 27 | 18 | 17 | 26 | 16 |
Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.
Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution.
Using the differences data, calculate the sample mean and the sample standard deviation. = 3.71, = 4.5.
Random variable: = mean difference in the distances between the hands.
Distribution for the hypothesis test: t6
H0: μd = 0 Ha: μd ≠ 0
Graph:
Calculate the p-value: The p-value is 0.0716 (using the data directly).
(test statistic = 2.18. p-value = 0.0719 using
Decision: Assume α = 0.05. Since α < p-value, Do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put.
Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.
| Player 1 | Player 2 | Player 3 | Player 4 | Player 5 | |
|---|---|---|---|---|---|
| Dominant Hand | 120 | 111 | 135 | 140 | 125 |
| Off-hand | 105 | 109 | 98 | 111 | 99 |
The p-level is 0.0230, so we can reject the null hypothesis. The data show that the players do not throw the same distance with their off-hands as they do with their dominant hands.
Class Time:
Names:
Student Learning Outcomes
Increasing Stocks Survey
Decreasing Stocks Survey
Candy Survey Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands.
Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample.
Two population means from independent samples where the population standard deviations are not known
A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations), will have these characteristics:
Test of two population proportions from independent samples.
A hypothesis test for matched or paired samples (t-test) has these characteristics:
Standard error: SE =
Test statistic (t-score): t =
Degrees of freedom:
where:
s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
and are the sample means.
Cohen’s d is the measure of effect size:
Normal Distribution:
Test Statistic (z-score):
where:
Pooled Proportion: pc =
Distribution for the differences:
where the null hypothesis is H0: pA = pB or H0: pA – pB = 0.
Test Statistic (z-score):
where the null hypothesis is H0: pA = pB or H0: pA − pB = 0.
where
p′A and p′B are the sample proportions, pA and pB are the population proportions,
Pc is the pooled proportion, and nA and nB are the sample sizes.
Test Statistic (t-score): t =
where:
is the mean of the sample differences. μd is the mean of the population differences. sd is the sample standard deviation of the differences. n is the sample size.
Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for
It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal.
A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this.
A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.
The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their mid-level professionals are paid differently, on average.
The average worker in Germany gets eight weeks of paid vacation.
According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market.
It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four.
The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different?
In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico?
A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after.
It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.
Varsity athletes practice five times a week, on average.
A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different?
A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job?
A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes?
Are standard deviations known or unknown?
What is the random variable?
Is this a one-tailed or two-tailed test?
Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.
Is this a test of means or proportions?
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
In symbols, what is the random variable of interest for this test?
In words, define the random variable of interest for this test.
Which distribution (normal or Student's t) would you use for this hypothesis test?
Explain why you chose the distribution you did for Exercise 10.24.
Calculate the test statistic and p-value.
Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
Find the p-value.
At a pre-conceived α = 0.05, what is your:
Does it appear that the means are the same? Why or why not?
Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball.
| Pitcher | Sample Mean Speed of Pitches (mph) | Population Standard Deviation |
|---|---|---|
| Wesley | 86 | 3 |
| Rodriguez | 91 | 7 |
What is the random variable?
State the null and alternative hypotheses.
What is the test statistic?
What is the p-value?
At the 1% significance level, what is your conclusion?
Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.
| Plant Group | Sample Mean Height of Plants (inches) | Population Standard Deviation |
|---|---|---|
| Food | 16 | 2.5 |
| No food | 14 | 1.5 |
Is the population standard deviation known or unknown?
State the null and alternative hypotheses.
What is the p-value?
Draw the graph of the p-value.
At the 1% significance level, what is your conclusion?
| Sample Mean Melting Temperatures (°F) | Population Standard Deviation | |
|---|---|---|
| Alloy Gamma | 800 | 95 |
| Alloy Zeta | 900 | 105 |
State the null and alternative hypotheses.
Is this a right-, left-, or two-tailed test?
What is the p-value?
Draw the graph of the p-value.
At the 1% significance level, what is your conclusion?
Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.
Is this a test of means or proportions?
What is the random variable?
State the null and alternative hypotheses.
What is the p-value?
What can you conclude about the two operating systems?
Is this a test of means or proportions?
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
What is the random variable of interest for this test?
In words, define the random variable for this test.
Which distribution (normal or Student's t) would you use for this hypothesis test?
Explain why you chose the distribution you did for the Exercise 10.56.
Calculate the test statistic.
Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
Find the p-value.
At a pre-conceived α = 0.05, what is your:
Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?
Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.
| Installation | A | B | C | D | E | F | G | H |
|---|---|---|---|---|---|---|---|---|
| Before | 3 | 6 | 4 | 2 | 5 | 8 | 2 | 6 |
| After | 1 | 5 | 2 | 0 | 1 | 0 | 2 | 2 |
What is the random variable?
State the null and alternative hypotheses.
What is the p-value?
Draw the graph of the p-value.
What conclusion can you draw about the software patch?
| Subject | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| Before | 3 | 4 | 3 | 2 | 4 | 5 |
| After | 4 | 5 | 6 | 4 | 5 | 7 |
State the null and alternative hypotheses.
What is the p-value?
What is the sample mean difference?
Draw the graph of the p-value.
What conclusion can you draw about the juggling class?
| Patient | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| Before | 161 | 162 | 165 | 162 | 166 | 171 |
| After | 158 | 159 | 166 | 160 | 167 | 169 |
State the null and alternative hypotheses.
What is the test statistic?
What is the p-value?
What is the sample mean difference?
What is the conclusion?
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
If you are using a Student's t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)
The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same?
A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.
At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis.
| Relaxed time (seconds) | Jumping time (seconds) |
|---|---|
| 26 | 21 |
| 47 | 40 |
| 30 | 28 |
| 22 | 21 |
| 23 | 25 |
| 45 | 43 |
| 37 | 35 |
| 29 | 32 |
Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary.
Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean.
Use the information from Table 13.45 to answer the next four exercises.
Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices.
Repeat the test in Exercise 10.83, but use Lap 5 data this time.
Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.
In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?”
Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.
| Western | Eastern |
|---|---|
| Los Angeles 9 | D.C. United 9 |
| FC Dallas 3 | Chicago 8 |
| Chivas USA 4 | Columbus 7 |
| Real Salt Lake 3 | New England 6 |
| Colorado 4 | MetroStars 5 |
| San Jose 4 | Kansas City 3 |
Conduct a hypothesis test to answer the next two exercises.
The exact distribution for the hypothesis test is:
If the level of significance is 0.05, the conclusion is:
Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is:
Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level.
A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds.
Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is:
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Table 13.45. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)
A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3.
Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls.
A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same.
Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim.
A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test.
One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). Table 10.26 contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).
| Wife’s Score | 2 | 2 | 3 | 3 | 4 | 2 | 1 | 1 | 2 | 4 |
| Husband’s Score | 2 | 2 | 1 | 3 | 2 | 1 | 1 | 1 | 2 | 4 |
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Table 13.45. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them.
We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the same for the whites and the blacks races in the United States. We randomly pick one year, 1992, to compare the races. The number of suicides estimated in the United States in 1992 for white females is 4,930. Five hundred eighty were aged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let female suicide victims be our population.
Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?
A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different?
Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test.
This is:
An appropriate null hypothesis is:
The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is
Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders.
Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table 10.27. Test at the 1% level of significance.
| Number who are obese | Sample size | |
|---|---|---|
| Men | 42,769 | 155,525 |
| Women | 67,169 | 248,775 |
Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1% level of significance.
| 16–29 year olds | 30 years old and older | |
|---|---|---|
| Own a Tablet | 69 | 231 |
| Sample Size | 628 | 2,309 |
A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5% level of significance.
While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.
We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software.
Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe that the proportion of males has reached the proportion of females?
Use the data sets found in Table 13.45 to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds?
"To Breakfast or Not to Breakfast?" by Richard Ayore
In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come.
If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine.
Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!”
And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table 10.29, solve our problem.
| Work hours with breakfast | Work hours without breakfast |
|---|---|
| 8 | 6 |
| 7 | 5 |
| 9 | 5 |
| 5 | 4 |
| 9 | 7 |
| 8 | 7 |
| 10 | 7 |
| 7 | 5 |
| 6 | 6 |
| 9 | 5 |
DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
If you are using a Student's t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)
Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered?
| Starting cholesterol level | Ending cholesterol level |
|---|---|
| 140 | 140 |
| 220 | 230 |
| 110 | 120 |
| 240 | 220 |
| 200 | 190 |
| 180 | 150 |
| 190 | 200 |
| 360 | 300 |
| 280 | 300 |
| 260 | 240 |
Use the following information to answer the next two exercises. A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same.
Let the subscript t = treated patient and ut = untreated patient.
The appropriate hypotheses are:
If the p-value is 0.0062 what is the conclusion (use α = 0.05)?
Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: = −10.2 sd = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training.
The distribution for the test is:
If α = 0.05, the p-value and the conclusion are
A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows.
| Player 1 | Player 2 | Player 3 | Player 4 | |
|---|---|---|---|---|
| Mean score before class | 83 | 78 | 93 | 87 |
| Mean score after class | 80 | 80 | 86 | 86 |
The correct decision is:
A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in Table 10.32.
| Southern States | 2012 | 2013 |
|---|---|---|
| Alabama | 3,450 | 3,720 |
| Arkansas | 2,150 | 2,280 |
| Florida | 15,540 | 15,710 |
| Georgia | 6,970 | 7,310 |
| Kentucky | 3,160 | 3,300 |
| Louisiana | 3,320 | 3,630 |
| Mississippi | 1,990 | 2,080 |
| North Carolina | 7,090 | 7,430 |
| Oklahoma | 2,630 | 2,690 |
| South Carolina | 3,570 | 3,580 |
| Tennessee | 4,680 | 5,070 |
| Texas | 15,050 | 14,980 |
| Virginia | 6,190 | 6,280 |
A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in Table 10.33. Test at the 1% level of significance.
| Cities | Hyatt Regency prices in dollars | Hilton prices in dollars |
|---|---|---|
| Atlanta | 107 | 169 |
| Boston | 358 | 289 |
| Chicago | 209 | 299 |
| Dallas | 209 | 198 |
| Denver | 167 | 169 |
| Indianapolis | 179 | 214 |
| Los Angeles | 179 | 169 |
| New York City | 625 | 459 |
| Philadelphia | 179 | 159 |
| Washington, DC | 245 | 239 |
A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in Table 10.34.
| Northeastern States | 2011 | 2012 |
|---|---|---|
| Connecticut | 17.3 | 16.4 |
| Delaware | 17.4 | 13.7 |
| Maine | 19.3 | 16.1 |
| Maryland | 16.0 | 15.5 |
| Massachusetts | 17.6 | 18.2 |
| New Hampshire | 15.4 | 13.5 |
| New Jersey | 19.2 | 18.7 |
| New York | 18.5 | 18.7 |
| Ohio | 18.2 | 18.8 |
| Pennsylvania | 16.5 | 16.9 |
| Rhode Island | 20.7 | 22.4 |
| Vermont | 14.7 | 12.3 |
| West Virginia | 15.5 | 17.3 |
Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test.
A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.
A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.
The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females.
A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.
A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.
According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this.
According to a recent study, U.S. companies have a mean maternity-leave of six weeks.
A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally.
A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected:
| Pre-course score | Post-course score |
|---|---|
| 1 | 300 |
| 960 | 920 |
| 1010 | 1100 |
| 840 | 880 |
| 1100 | 1070 |
| 1250 | 1320 |
| 860 | 860 |
| 1330 | 1370 |
| 790 | 770 |
| 990 | 1040 |
| 1110 | 1200 |
| 740 | 850 |
University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.
Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown Table 10.36. Determine the appropriate test and best distribution to use for that test.
| Left-handed | Right-handed | |
| Sample size | 41 | 41 |
| Sample mean | 97.5 | 98.1 |
| Sample standard deviation | 17.5 | 19.2 |
A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table 10.37.
| Player 1 | Player 2 | Player 3 | Player 4 | |
|---|---|---|---|---|
| Mean score before class | 83 | 78 | 93 | 87 |
| Mean score after class | 80 | 80 | 86 | 86 |
This is:
Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com
Data from Microsoft Bookshelf.
Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).
“List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age (accessed June 17, 2013).
“Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013).
“Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).
“World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).
Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf
Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013).
“Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013).
Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013).
“State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013).
“Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013).
Data from Educational Resources, December catalog.
Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013).
Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013).
Data from Statistics, United States Department of Health and Human Services.
Data from Whitney Exhibit on loan to San Jose Museum of Art.
Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013).
Data from the Chancellor’s Office, California Community Colleges, November 1994.
“State of the States.” Gallup, 2013. Available online at http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive (accessed June 17, 2013).
“West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/westnile/index.htm (accessed June 17, 2013).
two proportions
matched or paired samples
single mean
independent group means, population standard deviations and/or variances unknown
two proportions
independent group means, population standard deviations and/or variances unknown
independent group means, population standard deviations and/or variances unknown
two proportions
The random variable is the difference between the mean amounts of sugar in the two soft drinks.
means
two-tailed
the difference between the mean life spans of whites and nonwhites
This is a comparison of two population means with unknown population standard deviations.
Check student’s solution.
The difference in mean speeds of the fastball pitches of the two pitchers
–2.46
At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s.
0.0062
There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.
P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2.
0.1018
proportions
right-tailed
The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.
Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.
Check student’s solution.
the mean difference of the system failures
0.0067
With a p-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.
0.0021
0.0699
We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.
Subscripts: 1: two-year colleges; 2: four-year colleges
Subscripts: 1: mechanical engineering; 2: electrical engineering
c
Test: two independent sample means, population standard deviations unknown.
Random variable:
Distribution: H0: μ1 = μ2 Ha: μ1 < μ2 The mean age of entering prostitution in Canada is lower than the mean age in the United States.
Graph: left-tailed
p-value : 0.0151
Decision: Do not reject H0.
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States.
d
Subscripts: 1 = boys, 2 = girls
Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans
Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College
a
Test: two independent sample proportions.
Random variable: p′1 - p′2
Distribution:
The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.
Graph: two-tailed
p-value : 0.0033
Decision: Reject the null hypothesis.
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older.
Test: two independent sample proportions
Random variable: p′1 − p′2
Distribution:
H0: p1 = p2
A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.
Graph: right-tailed
Decision: Do not reject the H0.
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.
Subscripts: 1: men; 2: women
p-value = 0.1494
At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.
b
c
Test: two matched pairs or paired samples (t-test)
Random variable:
Distribution: t12
H0: μd = 0 Ha: μd > 0
The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.
Graph: right-tailed
p-value: 0.0004
Decision: Reject H0
Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.
Test: matched or paired samples (t-test)
Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8}
Random Variable:
Distribution: H0: μd = 0 Ha: μd < 0
The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.
Graph: left-tailed.
p-value: 0.1207
Decision: Do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.
e
d
f
e
f
a
By the end of this chapter, the student should be able to:
Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test.
You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution.
In this chapter, you will learn the three major applications of the chi-square distribution:
Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see Notes for the TI-83, 83+, 84, 84+ Calculators). TI-83+ and TI-84 calculator instructions are included in the text.
Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like). Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice.
The notation for the chi-square distribution is:
For the χ2 distribution, the population mean is μ = df and the population standard deviation is .
The random variable is shown as χ2, but may be any upper case letter.
The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables.
χ2 = (Z1)2 + (Z2)2 + ... + (Zk)2
In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.
The test statistic for a goodness-of-fit test is:
where:
The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form .
The number of degrees of freedom is df = (number of categories – 1).
The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.
The expected value for each cell needs to be at least five in order for you to use this test.
Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to Table 11.1.
| Number of absences per term | Expected number of students |
|---|---|
| 0–2 | 50 |
| 3–5 | 30 |
| 6–8 | 12 |
| 9–11 | 6 |
| 12+ | 2 |
A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in Table 11.2 displays the results of that survey.
| Number of absences per term | Actual number of students |
|---|---|
| 0–2 | 35 |
| 3–5 | 40 |
| 6–8 | 20 |
| 9–11 | 1 |
| 12+ | 4 |
H0: Student absenteeism fits faculty perception.
Ha: Student absenteeism does not fit faculty perception.
a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?
a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are in Table 11.3 and Table 11.4.
| Number of absences per term | Expected number of students |
|---|---|
| 0–2 | 50 |
| 3–5 | 30 |
| 6–8 | 12 |
| 9+ | 8 |
| Number of absences per term | Actual number of students |
|---|---|
| 0–2 | 35 |
| 3–5 | 40 |
| 6–8 | 20 |
| 9+ | 5 |
b. What is the number of degrees of freedom (df)?
b. There are four "cells" or categories in each of the new tables.
df = number of cells – 1 = 4 – 1 = 3
A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5.
| Number produced | Number defective |
|---|---|
| 0–100 | 5 |
| 101–200 | 6 |
| 201–300 | 7 |
| 301–400 | 8 |
| 401–500 | 10 |
A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey.
| Number produced | Number defective |
|---|---|
| 0–100 | 5 |
| 101–200 | 7 |
| 201–300 | 8 |
| 301–400 | 9 |
| 401–500 | 11 |
State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.
H0:The number of defaults fits expectations.
Ha:The number of defaults does not fit expectations.
Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table 11.7. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.
| Monday | Tuesday | Wednesday | Thursday | Friday | |
|---|---|---|---|---|---|
| Number of Absences | 15 | 12 | 9 | 9 | 15 |
The null and alternative hypotheses are:
If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data.
This time, calculate the χ2 test statistic by hand. Make a chart with the following headings and fill in the columns:
Now add (sum) the last column. The sum is three. This is the χ2 test statistic.
To find the p-value, calculate P(χ2 > 3). This test is right-tailed. (Use a computer or calculator to find the p-value. You should get p-value = 0.5578.)
The dfs are the number of cells – 1 = 5 – 1 = 4
Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the p-value.
Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)
The decision is not to reject the null hypothesis.
Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.
TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example Example 11.2 has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER.
Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in Table 11.8.
| Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | |
|---|---|---|---|---|---|---|---|
| Number of Students | 11 | 8 | 10 | 7 | 10 | 5 | 5 |
From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?
df = 6
p-value = 0.6093
One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in Table 11.9.
| Number of Televisions | Percent |
|---|---|
| 0 | 10 |
| 1 | 16 |
| 2 | 55 |
| 3 | 11 |
| 4+ | 8 |
The table contains expected (E) percents.
A random sample of 600 families in the far western United States resulted in the data in Table 11.10.
| Number of Televisions | Frequency |
|---|---|
| Total = 600 | |
| 0 | 66 |
| 1 | 119 |
| 2 | 340 |
| 3 | 60 |
| 4+ | 15 |
The table contains observed (O) frequency values.
At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole?
This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.
The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in Table 11.11.
| Number of Televisions | Percent | Expected Frequency |
|---|---|---|
| 0 | 10 | (0.10)(600) = 60 |
| 1 | 16 | (0.16)(600) = 96 |
| 2 | 55 | (0.55)(600) = 330 |
| 3 | 11 | (0.11)(600) = 66 |
| over 3 | 8 | (0.08)(600) = 48 |
Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.
H0: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.
Ha: The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.
Distribution for the test: where df = (the number of cells) – 1 = 5 – 1 = 4.
df ≠ 600 – 1
Calculate the test statistic: χ2 = 29.65
Graph:
Probability statement: p-value = P(χ2 > 29.65) = 0.000006
Compare α and the p-value:
Make a decision: Since α > p-value, reject Ho.
This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.
Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole.
Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example 11.1). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.
The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12.
| Number of Pets | Percent |
|---|---|
| 0 | 18 |
| 1 | 25 |
| 2 | 30 |
| 3 | 18 |
| 4+ | 9 |
A random sample of 1,000 students from the Eastern United States resulted in the data in Table 11.13.
| Number of Pets | Frequency |
|---|---|
| 0 | 210 |
| 1 | 240 |
| 2 | 320 |
| 3 | 140 |
| 4+ | 90 |
At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?
p-value = 0.0036
We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole.
Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level.
This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?"
Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.
H0: The coins are fair.
Ha: The coins are not fair.
Distribution for the test: where df = 3 – 1 = 2.
Calculate the test statistic: χ2 = 2.14
Graph:
Probability statement: p-value = P(χ2 > 2.14) = 0.3430
Compare α and the p-value:
Make a decision: Since α < p-value, do not reject H0.
Conclusion: There is insufficient evidence to conclude that the coins are not fair.
Press STAT and ENTER. Make sure you
clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed
frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow
over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and
ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press
5. You should see "sum".Enter L3. Rounded to two decimal places, you
should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press
ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which
is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make
sure you clear any lists before you start.
Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. Table 11.14 shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?
| MDG Region | Adult Literacy Rate (%) |
|---|---|
| Developed Regions | 99.0 |
| Commonwealth of Independent States | 99.5 |
| Northern Africa | 67.3 |
| Sub-Saharan Africa | 62.5 |
| Latin America and the Caribbean | 91.0 |
| Eastern Asia | 93.8 |
| Southern Asia | 61.9 |
| South-Eastern Asia | 91.9 |
| Western Asia | 84.5 |
| Oceania | 66.4 |
degrees of freedom = 9
chi2 test statistic = 26.38
Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4. Into L2, put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum". Enter L3. Rounded to two decimal places, you should see 26.38. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 26.38,1E99,9). Rounded to four places, you should see .0018, which is the p-value.
The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.
Tests of independence involve using a contingency table of observed (data) values.
The test statistic for a test of independence is similar to that of a goodness-of-fit test:
where:
There are terms of the form .
A test of independence determines whether two factors are independent or not. You first encountered the term independence in Probability Topics. As a review, consider the following example.
The expected value for each cell needs to be at least five in order for you to use this test.
Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P(A AND B) = P(A)P(B). A AND B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not.
Let y = expected number of drivers who used a cell phone while driving and received speeding violations.
If A and B are independent, then P(A AND B) = P(A)P(B). By substitution,
Solve for y: y =
About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.
In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is:
H0: Being a cell phone user while driving and receiving a speeding violation are independent events.
If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation.
The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.
The number of degrees of freedom for the test of independence is:
df = (number of columns - 1)(number of rows - 1)
The following formula calculates the expected number (E):
A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll?
About 16 students are expected to be music students and on the honor roll.
In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table 11.15 is a sample of the adult volunteers and the number of hours they volunteer per week.
| Type of Volunteer | 1–3 Hours | 4–6 Hours | 7–9 Hours | Row Total |
|---|---|---|---|---|
| Community College Students | 111 | 96 | 48 | 255 |
| Four-Year College Students | 96 | 133 | 61 | 290 |
| Nonstudents | 91 | 150 | 53 | 294 |
| Column Total | 298 | 379 | 162 | 839 |
Is the number of hours volunteered independent of the type of volunteer?
The observed table and the question at the end of the problem, "Is the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed.
H0: The number of hours volunteered is independent of the type of volunteer.
Ha: The number of hours volunteered is dependent on the type of volunteer.
The expected result are in Table 11.16.
| Type of Volunteer | 1-3 Hours | 4-6 Hours | 7-9 Hours |
|---|---|---|---|
| Community College Students | 90.57 | 115.19 | 49.24 |
| Four-Year College Students | 103.00 | 131.00 | 56.00 |
| Nonstudents | 104.42 | 132.81 | 56.77 |
For example, the calculation for the expected frequency for the top left cell is
Calculate the test statistic: χ2 = 12.99 (calculator or computer)
Distribution for the test:
df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4
Graph:
Probability statement: p-value=P(χ2 > 12.99) = 0.0113
Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0113. α > p-value.
Make a decision: Since α > p-value, reject H0. This means that the factors are not independent.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.
For the example in Table 11.16, if there had been another type of volunteer, teenagers, what would the degrees of freedom be?
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Table 11.16. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate.
The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table 11.17 shows the results:
| Industry Sector | 2000 | 2010 | 2020 | Total |
|---|---|---|---|---|
| Nonagriculture wage and salary | 13,243 | 13,044 | 15,018 | 41,305 |
| Goods-producing, excluding agriculture | 2,457 | 1,771 | 1,950 | 6,178 |
| Services-providing | 10,786 | 11,273 | 13,068 | 35,127 |
| Agriculture, forestry, fishing, and hunting | 240 | 214 | 201 | 655 |
| Nonagriculture self-employed and unpaid family worker | 931 | 894 | 972 | 2,797 |
| Secondary wage and salary jobs in agriculture and private household industries | 14 | 11 | 11 | 36 |
| Secondary jobs as a self-employed or unpaid family worker | 196 | 144 | 152 | 492 |
| Total | 27,867 | 27,351 | 31,372 | 86,590 |
We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom.
H0 : The number of jobs is independent of the year.
Ha : The number of jobs is dependent on the year.
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 227.73 and the p−value = 5.90E - 42 = 0. Do the procedure a second time but arrow down to Draw instead of calculate.
De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table 11.18 shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.
| Need to Succeed in School | High |
Med-high |
Medium |
Med-low |
Low |
Row Total |
|---|---|---|---|---|---|---|
| High Need | 35 | 42 | 53 | 15 | 10 | 155 |
| Medium Need | 18 | 48 | 63 | 33 | 31 | 193 |
| Low Need | 4 | 5 | 11 | 15 | 17 | 52 |
| Column Total | 57 | 95 | 127 | 63 | 58 | 400 |
a. How many high anxiety level students are expected to have a high need to succeed in school?
a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.
The expected number of students who have a high anxiety level and a high need to succeed in school is about 22.
b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?
b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.
c. = ________
c.
d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.
d. 8
Refer back to the information in Try It. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?
12,727, 14,965
The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.
The expected value for each cell needs to be at least five in order for you to use this test.
Hypotheses
Test StatisticUse a test statistic. It is computed in the same way as the test for independence.
Degrees of Freedom (df)df = number of columns - 1
RequirementsAll values in the table must be greater than or equal to five.
Common UsesComparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.
Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 11.19. Do male and female college students have the same distribution of living arrangements?
| Dormitory | Apartment | With Parents | Other | |
| Males | 72 | 84 | 49 | 45 |
| Females | 91 | 86 | 88 | 35 |
H0: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students.
MATRXkey and arrow over to
EDIT. Press
1:[A]. Press
2 ENTER 4 ENTER. Enter the table values by row. Press
ENTERafter each. Press
2nd QUIT. Press
STATand arrow over to
TESTS. Arrow down to
C:χ2-TEST. Press
ENTER. You should see
Observed:[A] and Expected:[B]. Arrow down to
Calculate. Press
ENTER. The test statistic is 10.1287 and the p-value = 0.0175. Do the procedure a second time but arrow down to
Drawinstead of
calculate.
Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 11.20. Do families and singles have the same distribution of cars? Test at a level of significance of 0.05.
| Sport | Sedan | Hatchback | Truck | Van/SUV | |
|---|---|---|---|---|---|
| Family | 5 | 15 | 35 | 17 | 28 |
| Single | 45 | 65 | 37 | 46 | 7 |
With a p-value of almost zero, we reject the null hypothesis. The data show that the distribution of cars is not the same for families and singles.
Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 11.21 shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake?
| Perez | Chung | Stevens | |
| Before | 167 | 128 | 135 |
| After | 214 | 197 | 225 |
H0: The distribution of voter preferences was the same before and after the earthquake.
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 3.2603 and the p-value = 0.1959. Do the procedure a second time but arrow down to Draw instead of calculate.
Compare α and the p-value: α = 0.05 and the p-value = 0.1959. α < p-value.
Make a decision: Since α < p-value, do not reject Ho.
Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake.
Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 11.22, two types of applications are accepted: regular and early decision.
| Application Type Accepted | Brown | Columbia | Cornell | Dartmouth | Penn | Yale |
|---|---|---|---|---|---|---|
| Regular | 2,115 | 1,792 | 5,306 | 1,734 | 2,685 | 1,245 |
| Early Decision | 577 | 627 | 1,228 | 444 | 1,195 | 761 |
We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity.
H0 : The distribution of regular applications accepted is the same as the distribution of early applications accepted.
Ha : The distribution of regular applications accepted is not the same as the distribution of early applications accepted.
Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down toC:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 430.06 and the p-value = 9.80E-91. Do the procedure a second time but arrow down to Draw instead of calculate.
You have seen the χ2 test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which χ2 test is the appropriate one to use.
A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is:
where:
You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.9 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.
Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average.
Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?
Even though we are given the population standard deviation, we can set up the test using the population variance as follows.
A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?
H0: σ2 = 32
Ha: σ2 < 32
With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.
With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers.
Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ.
Random Variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times.
The word "less" tells you this is a left-tailed test.
Distribution for the test: , where:
Calculate the test statistic:
where n = 25, s = 3.5, and σ = 7.2.
Graph:
Probability statement: p-value = P ( χ2 < 5.67) = 0.000042
Compare α and the p-value:
Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.
In 2nd DISTR, use 7:χ2cdf. The syntax is
(lower, upper, df) for the parameter list.
For Example 11.10, χ2cdf(-1E99,5.67,24). The p-value = 0.000042.
The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the 1% significance level.
H0: σ2 = 12.22
Ha: σ2 > 12.22
The p-value is 0.2902, so we decline to reject the null hypothesis. There is not enough evidence to suggest that the variance is greater than 12.22.
In 2nd DISTR, use7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. χ2cdf(16.39,10^99,14). The p-value = 0.2902.
Class Time:
Names:
Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. (Or, ask three cashiers for the last ten amounts. Be sure to include the express lane, if it is open.)
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
| __________ | __________ | __________ | __________ | __________ |
Uniform Distribution Test to see if grocery receipts follow the uniform distribution.
| Fifth | Observed | Expected |
|---|---|---|
| 1st | ||
| 2nd | ||
| 3rd | ||
| 4th | ||
| 5th |
Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter .
| Cell | Observed | Expected |
|---|---|---|
| 1st | ||
| 2nd | ||
| 3rd | ||
| 4th | ||
| 5th | ||
| 6th |
Class Time:
Names:
| sweets (candy & baked goods) | ice cream | chips & pretzels | fruits & vegetables | Total | |
|---|---|---|---|---|---|
| male | |||||
| female | |||||
| Total |
Hypothesis Test Conduct a hypothesis test to determine if the factors are independent:
The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population.
An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom.
The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests.
To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.
To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5.
To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five.
The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown.
To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).
χ2 = (Z1)2 + (Z2)2 + … (Zdf)2 chi-square distribution random variable
μχ2 = df chi-square distribution population mean
Chi-Square distribution population standard deviation
goodness-of-fit test statistic where:
df = k − 1 degrees of freedom
Homogeneity test statistic where: O = observed values
df = (i −1)(j −1) Degrees of freedom
Test of a single variance statistic where:
df = n – 1 Degrees of freedom
If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation?
If df > 90, the distribution is _____________. If df = 15, the distribution is ________________.
When does the chi-square curve approximate a normal distribution?
Where is μ located on a chi-square curve?
Is it more likely the df is 90, 20, or two in the graph?
Determine the appropriate test to be used in the next three exercises.
An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.
An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.
A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.
Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27.
| Grade | Proportion |
|---|---|
| A | 0.25 |
| B | 0.30 |
| C | 0.35 |
| D | 0.10 |
The actual distribution for a class of 20 is in Table 11.28.
| Grade | Frequency |
|---|---|
| A | 7 |
| B | 7 |
| C | 5 |
| D | 1 |
______
State the null and alternative hypotheses.
χ2 test statistic = ______
p-value = ______
At the 5% significance level, what can you conclude?
| Ethnicity | Number of Cases |
|---|---|
| White | 2,229 |
| Hispanic | 1,157 |
| Black/African-American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
The percentage of each ethnic group in Santa Clara County is as in Table 11.30.
| Ethnicity | Percentage of total county population | Number expected (round to two decimal places) |
|---|---|---|
| White | 42.9% | 1748.18 |
| Hispanic | 26.7% | |
| Black/African-American | 2.6% | |
| Asian, Pacific Islander | 27.8% | |
| Total = 100% |
If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
H0: _______
Ha: _______
Is this a right-tailed, left-tailed, or two-tailed test?
degrees of freedom = _______
χ2 test statistic = _______
p-value = _______
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.
Let α = 0.05
Decision: ________________
Reason for the Decision: ________________
Conclusion (write out in complete sentences): ________________
Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?
Determine the appropriate test to be used in the next three exercises.
A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups.
The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. He takes a random sample of 100 players from different organizations.
A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. She takes a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing.
| Traveling Distance | Third class | Second class | First class | Total |
|---|---|---|---|---|
| 1–100 miles | 21 | 14 | 6 | 41 |
| 101–200 miles | 18 | 16 | 8 | 42 |
| 201–300 miles | 16 | 17 | 15 | 48 |
| 301–400 miles | 12 | 14 | 21 | 47 |
| 401–500 miles | 6 | 6 | 10 | 22 |
| Total | 73 | 67 | 60 | 200 |
State the hypotheses.
df = _______
How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets?
How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets?
What is the test statistic?
What is the p-value?
What can you conclude at the 5% level of significance?
Complete the table.
| Smoking Level Per Day | African American | Native Hawaiian | Latino | Japanese Americans | White | TOTALS |
|---|---|---|---|---|---|---|
| 1-10 | ||||||
| 11-20 | ||||||
| 21-30 | ||||||
| 31+ | ||||||
| TOTALS |
State the hypotheses.
Enter expected values in Table 11.32. Round to two decimal places.
Calculate the following values:
df = _______
test statistic = ______
p-value = ______
Is this a right-tailed, left-tailed, or two-tailed test? Explain why.
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.
State the decision and conclusion (in a complete sentence) for the following preconceived levels of α.
α = 0.05
α = 0.01
A math teacher wants to see if two of her classes have the same distribution of test scores. What test should she use?
What are the null and alternative hypotheses for Exercise 11.43?
A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should he use?
A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should she use?
What condition must be met to use the test for homogeneity?
Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33.
| 20–30 | 30–40 | 40–50 | 50–60 | |
|---|---|---|---|---|
| Private Practice | 16 | 40 | 38 | 6 |
| Hospital | 8 | 44 | 59 | 39 |
State the null and alternative hypotheses.
df = _______
What is the test statistic?
What is the p-value?
What can you conclude at the 5% significance level?
Which test do you use to decide whether an observed distribution is the same as an expected distribution?
What is the null hypothesis for the type of test from Exercise 11.53?
Which test would you use to decide whether two factors have a relationship?
Which test would you use to decide if two populations have the same distribution?
How are tests of independence similar to tests for homogeneity?
How are tests of independence different from tests for homogeneity?
Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less.
What type of test should be used?
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
What type of test should be used?
State the null and alternative hypotheses.
df = ________
What type of test should be used?
What is the test statistic?
What is the p-value?
What can you conclude at the 5% significance level?
Decide whether the following statements are true or false.
As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical.
The standard deviation of the chi-square distribution is twice the mean.
The mean and the median of the chi-square distribution are the same if df = 24.
For each problem, use a solution sheet to solve the hypothesis test problem. Go to Table 13.45 for the chi-square solution sheet. Round expected frequency to two decimal places.
A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in Table 11.34 are the result of the 120 rolls.
| Face Value | Frequency | Expected Frequency |
|---|---|---|
| 1 | 15 | |
| 2 | 29 | |
| 3 | 16 | |
| 4 | 15 | |
| 5 | 30 | |
| 6 | 15 |
The marital status distribution of the U.S. male population, ages 15 and older, is as shown in Table 11.35.
| Marital Status | Percent | Expected Frequency |
|---|---|---|
| never married | 31.3 | |
| married | 56.1 | |
| widowed | 2.5 | |
| divorced/separated | 10.1 |
Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.36, rounding to two decimal places.
| Marital Status | Frequency |
|---|---|
| never married | 140 |
| married | 238 |
| widowed | 2 |
| divorced/separated | 20 |
Use the following information to answer the next two exercises: The columns in Table 11.37 contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam.
| Race/Ethnicity | AP Examinee Population | Overall Student Population | Survey Frequency |
|---|---|---|---|
| Asian, Asian American, or Pacific Islander | 10.2% | 5.4% | 113 |
| Black or African-American | 8.2% | 14.5% | 94 |
| Hispanic or Latino | 15.5% | 15.9% | 136 |
| American Indian or Alaska Native | 0.6% | 1.2% | 10 |
| White | 59.4% | 61.6% | 604 |
| Not reported/other | 6.1% | 1.4% | 43 |
Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.
Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.
The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in Table 11.38. Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area.
| Race | Lake Tahoe Frequency | Manhattan Frequency |
|---|---|---|
| Asian Indian | 131 | 174 |
| Chinese | 118 | 557 |
| Filipino | 1,045 | 518 |
| Japanese | 80 | 54 |
| Korean | 12 | 29 |
| Vietnamese | 9 | 21 |
| Other | 24 | 66 |
Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for Exercise 11.77 and Exercise 11.78. The second column in each table does not add to 100% because of rounding.
Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.
| Major | Women - Expected Major | Women - Actual Major |
|---|---|---|
| Arts & Humanities | 14.0% | 670 |
| Biological Sciences | 8.4% | 410 |
| Business | 13.1% | 685 |
| Education | 13.0% | 650 |
| Engineering | 2.6% | 145 |
| Physical Sciences | 2.6% | 125 |
| Professional | 18.9% | 975 |
| Social Sciences | 13.0% | 605 |
| Technical | 0.4% | 15 |
| Other | 5.8% | 300 |
| Undecided | 8.0% | 420 |
Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors.
| Major | Men - Expected Major | Men - Actual Major |
|---|---|---|
| Arts & Humanities | 11.0% | 600 |
| Biological Sciences | 6.7% | 330 |
| Business | 22.7% | 1130 |
| Education | 5.8% | 305 |
| Engineering | 15.6% | 800 |
| Physical Sciences | 3.6% | 175 |
| Professional | 9.3% | 460 |
| Social Sciences | 7.6% | 370 |
| Technical | 1.8% | 90 |
| Other | 8.2% | 400 |
| Undecided | 6.6% | 340 |
Read the statement and decide whether it is true or false.
In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true.
In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail.
Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not.
The test to use to determine if a six-sided die is fair is a goodness-of-fit test.
In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.
A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. Table 11.41 shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values.
| Business Type | Number in class | Observed Number that recycle one commodity | Expected number that recycle one commodity |
|---|---|---|---|
| Office | 35 | 19 | 17.5 |
| Retail/Wholesale | 48 | 27 | 24 |
| Food/Restaurants | 53 | 35 | 26.5 |
| Manufacturing/Medical | 52 | 21 | 26 |
| Hotel/Mixed | 24 | 9 | 12 |
Table 11.42 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population.
| Age Class (Years) | Obese (Percentage) | Expected USA average (Percentage) |
|---|---|---|
| 20–30 | 75.0 | 32.6 |
| 31–40 | 26.5 | 32.6 |
| 41–50 | 13.6 | 36.6 |
| 51–60 | 21.9 | 36.6 |
| 61–70 | 21.0 | 39.7 |
For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.
A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier.
| U.S. Ski Area | Beginner | Intermediate | Advanced |
|---|---|---|---|
| Tahoe | 20 | 30 | 40 |
| Utah | 10 | 30 | 60 |
| Colorado | 10 | 40 | 50 |
Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in Table 11.44. Conduct a test of independence.
| Family Size | Sub & Compact | Mid-size | Full-size | Van & Truck |
|---|---|---|---|---|
| 1 | 20 | 35 | 40 | 35 |
| 2 | 20 | 50 | 70 | 80 |
| 3–4 | 20 | 50 | 100 | 90 |
| 5+ | 20 | 30 | 70 | 70 |
College students may be interested in whether or not their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. Table 11.45 shows the data. Conduct a test of independence.
| Major | < $50,000 | $50,000 – $68,999 | $69,000 + |
|---|---|---|---|
| English | 5 | 20 | 5 |
| Engineering | 10 | 30 | 60 |
| Nursing | 10 | 15 | 15 |
| Business | 10 | 20 | 30 |
| Psychology | 20 | 30 | 20 |
Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in Table 11.46. Conduct a test of independence.
| Location | 20–29 | 30–39 | 40–49 | 50 and over |
|---|---|---|---|---|
| Niagara Falls | 15 | 25 | 25 | 20 |
| Poconos | 15 | 25 | 25 | 10 |
| Europe | 10 | 25 | 15 | 5 |
| Virgin Islands | 20 | 25 | 15 | 5 |
A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence.
| Sport | 18 - 25 | 26 - 30 | 31 - 40 | 41 and over |
|---|---|---|---|---|
| racquetball | 42 | 58 | 30 | 46 |
| tennis | 58 | 76 | 38 | 65 |
| swimming | 72 | 60 | 65 | 33 |
A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in Table 11.48. Conduct a test of independence.
| Type of Fries | Northeast | South | Central | West |
|---|---|---|---|---|
| skinny fries | 70 | 50 | 20 | 25 |
| curly fries | 100 | 60 | 15 | 30 |
| steak fries | 20 | 40 | 10 | 10 |
According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. He is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence.
| Age of Males | None | < $200,000 | $200,000–$400,000 | $401,001–$1,000,000 | $1,000,001+ |
|---|---|---|---|---|---|
| 20–29 | 40 | 15 | 40 | 0 | 5 |
| 30–39 | 35 | 5 | 20 | 20 | 10 |
| 40–49 | 20 | 0 | 30 | 0 | 30 |
| 50+ | 40 | 30 | 15 | 15 | 10 |
Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence.
| Annual Salary | Not a high school graduate | High school graduate | College graduate | Masters or doctorate |
|---|---|---|---|---|
| < $30,000 | 15 | 25 | 10 | 5 |
| $30,000–$40,000 | 20 | 40 | 70 | 30 |
| $40,000–$50,000 | 10 | 20 | 40 | 55 |
| $50,000–$60,000 | 5 | 10 | 20 | 60 |
| $60,000+ | 0 | 5 | 10 | 150 |
Read the statement and decide whether it is true or false.
The number of degrees of freedom for a test of independence is equal to the sample size minus one.
The test for independence uses tables of observed and expected data values.
The test to use when determining if the college or university a student chooses to attend is related to his or her socioeconomic status is a test for independence.
In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed.
An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on Table 11.51, do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5% significance level.
| U.S. region/Flavor | Strawberry | Chocolate | Vanilla | Rocky Road | Mint Chocolate Chip | Pistachio | Row total |
|---|---|---|---|---|---|---|---|
| West | 12 | 21 | 22 | 19 | 15 | 8 | 97 |
| Midwest | 10 | 32 | 22 | 11 | 15 | 6 | 96 |
| East | 8 | 31 | 27 | 8 | 15 | 7 | 96 |
| South | 15 | 28 | 30 | 8 | 15 | 6 | 102 |
| Column Total | 45 | 112 | 101 | 46 | 60 | 27 | 391 |
Table 11.52 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level.
| Age Group\ Net Worth Value (in millions of US dollars) | 1–5 | 6–24 | ≥25 | Row Total |
|---|---|---|---|---|
| 17–25 | 8 | 7 | 5 | 20 |
| 26–30 | 6 | 5 | 9 | 20 |
| Column Total | 14 | 12 | 14 | 40 |
A 2013 poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in Table 11.53, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5% significance level.
| Opinion/Ethnicity | Asian-American | White/Non-Hispanic | African-American | Latino | Row Total |
|---|---|---|---|---|---|
| Against tax | 48 | 433 | 41 | 160 | 628 |
| In Favor of tax | 54 | 234 | 24 | 147 | 459 |
| No opinion | 16 | 43 | 16 | 19 | 84 |
| Column Total | 118 | 710 | 71 | 272 | 1171 |
For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Table 13.45 for the chi-square solution sheet. Round expected frequency to two decimal places.
A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table 11.54. Conduct a test of homogeneity. Test at a 5% level of significance.
| Open | Conscientious | Extrovert | Agreeable | Neurotic | |
| Business | 41 | 52 | 46 | 61 | 58 |
| Social Science | 72 | 75 | 63 | 80 | 65 |
Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance.
| French Toast | Pancakes | Waffles | Omelettes | |
| Men | 47 | 35 | 28 | 53 |
| Women | 65 | 59 | 55 | 60 |
A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance.
In 2007, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In Table 11.56 you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5% significance level. Did you expect the result you obtained?
| Reasons for Homeschooling | Applicable Reason (in thousands of respondents) | Most Important Reason (in thousands of respondents) | Row Total |
|---|---|---|---|
| Concern about the environment of other schools | 1,321 | 309 | 1,630 |
| Dissatisfaction with academic instruction at other schools | 1,096 | 258 | 1,354 |
| To provide religious or moral instruction | 1,257 | 540 | 1,797 |
| Child has special needs, other than physical or mental | 315 | 55 | 370 |
| Nontraditional approach to child’s education | 984 | 99 | 1,083 |
| Other reasons (e.g., finances, travel, family time, etc.) | 485 | 216 | 701 |
| Column Total | 5,458 | 1,477 | 6,935 |
When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in Table 11.57 shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level.
| Year | European Union | United States | Row Total |
|---|---|---|---|
| 2010 | 3,413 | 7,164 | 10,557 |
| 2009 | 3,302 | 7,057 | 10,359 |
| 2008 | 3,505 | 7,488 | 10,993 |
| 2007 | 3,537 | 7,758 | 11,295 |
| 2006 | 3,595 | 7,697 | 11,292 |
| 2005 | 3,613 | 7,847 | 11,460 |
| Column Total | 45,011 | 20,965 | 65,976 |
The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. Table 11.58 presents the number of Top Safety Picks in six car categories for the two years 2009 and 2013. Analyze the table data to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between 2009 and 2013. Derive your results at the 5% significance level.
| Year \ Car Type | Small | Mid-Size | Large | Small SUV | Mid-Size SUV | Large SUV | Row Total |
|---|---|---|---|---|---|---|---|
| 2009 | 12 | 22 | 10 | 10 | 27 | 6 | 87 |
| 2013 | 31 | 30 | 19 | 11 | 29 | 4 | 124 |
| Column Total | 43 | 52 | 29 | 21 | 56 | 10 | 211 |
For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Table 13.45 for the chi-square solution sheet. Round expected frequency to two decimal places.
Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance.
Read the statement and decide whether it is true or false.
If df = 2, the chi-square distribution has a shape that reminds us of the exponential.
Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.
Is the traveler disputing the claim about the average or about the variance?
A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes.
Is this a right-tailed, left-tailed, or two-tailed test?
H0: __________
df = ________
chi-square test statistic = ________
p-value = ________
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value.
Let α = 0.05
How did you know to test the variance instead of the mean?
If an additional test were done on the claim of the average delay, which distribution would you use?
If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use?
For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Table 13.45 for the chi-square solution sheet. Round expected frequency to two decimal places.
A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated?
Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15.
Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes.
Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.
The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in Table 11.59. Does the students’ survey indicate that the standard deviation is greater than 0.75?
| # of births | Frequency |
|---|---|
| 0 | 5 |
| 1 | 30 |
| 2 | 10 |
| 3 | 5 |
According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11
The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz.
You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis?
A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test.
(a) at the 5% significance level
(b) at the 1% significance level
Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172;
Data from Parade Magazine.
“HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011.
Data from the U.S. Census Bureau
Data from the College Board. Available online at http://www.collegeboard.com.
Data from the U.S. Census Bureau, Current Population Reports.
Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92.
Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013).
Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013).
DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/Rls2436.pdf (accessed May 24, 2013).
Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-ice-cream (accessed May 24, 2013)
“Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013).
Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013).
“Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013).
“Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/pubsinfo.asp?pubid=2009030 (accessed May 24, 2013).
“Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/2009030_sup.pdf (accessed May 24, 2013).
“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).
Data from the World Bank, June 5, 2012.
mean = 25 and standard deviation = 7.0711
when the number of degrees of freedom is greater than 90
df = 2
a goodness-of-fit test
3
2.04
We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.
H0: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.
right-tailed
88,621
Graph: Check student’s solution.
Decision: Reject the null hypothesis.
Reason for the Decision: p-value < alpha
Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.
a test of independence
a test of independence
8
6.6
0.0435
| Smoking Level Per Day | African American | Native Hawaiian | Latino | Japanese Americans | White | Totals |
|---|---|---|---|---|---|---|
| 1-10 | 9,886 | 2,745 | 12,831 | 8,378 | 7,650 | 41,490 |
| 11-20 | 6,514 | 3,062 | 4,932 | 10,680 | 9,877 | 35,065 |
| 21-30 | 1,671 | 1,419 | 1,406 | 4,715 | 6,062 | 15,273 |
| 31+ | 759 | 788 | 800 | 2,305 | 3,970 | 8,622 |
| Totals | 18,830 | 8,014 | 19,969 | 26,078 | 27,559 | 10,0450 |
| Smoking Level Per Day | African American | Native Hawaiian | Latino | Japanese Americans | White |
|---|---|---|---|---|---|
| 1-10 | 7777.57 | 3310.11 | 8248.02 | 10771.29 | 11383.01 |
| 11-20 | 6573.16 | 2797.52 | 6970.76 | 9103.29 | 9620.27 |
| 21-30 | 2863.02 | 1218.49 | 3036.20 | 3965.05 | 4190.23 |
| 31+ | 1616.25 | 687.87 | 1714.01 | 2238.37 | 2365.49 |
10,301.8
right
test for homogeneity
test for homogeneity
All values in the table must be greater than or equal to five.
3
0.00005
a goodness-of-fit test
a test for independence
Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way . In addition, all values must be greater than or equal to five.
a test of a single variance
a left-tailed test
a test of a single variance
0.0542
true
false
| Marital Status | Percent | Expected Frequency |
|---|---|---|
| never married | 31.3 | 125.2 |
| married | 56.1 | 224.4 |
| widowed | 2.5 | 10 |
| divorced/separated | 10.1 | 40.4 |
true
true
false
Alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: p-value < alpha
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.
true
true
225
H0: σ2 ≤ 150
36
Check student’s solution.
The claim is that the variance is no more than 150 minutes.
a Student's t- or normal distribution
The sample standard deviation is $34.29.
H0 : σ2 = 252
By the end of this chapter, the student should be able to:
Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it?
In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee.
The type of data described in the examples is bivariate data — "bi" for two variables. In reality, statisticians use multivariate data, meaning many variables.
In this chapter, you will be studying the simplest form of regression, "linear regression" with one independent variable (x). This involves data that fits a line in two dimensions. You will also study correlation which measures how strong the relationship is.
Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form:
The variable x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable.
The following examples are linear equations.
Is the following an example of a linear equation?
y = –0.125 – 3.5x
yes
The graph of a linear equation of the form y = a + bx is a straight line. Any line that is not vertical can be described by this equation.
Graph the equation y = –1 + 2x.
Is the following an example of a linear equation? Why or why not?
No, the graph is not a straight line; therefore, it is not a linear equation.
Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job.
Find the equation that expresses the total cost in terms of the number of hours required to complete the job.
Let x = the number of hours it takes to get the job done.
The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)(x) is the cost of the word processing only. The total cost is: y = 31.50 + 32x
Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class.
y = 50 + 20x
For the linear equation y = a + bx, b = slope and a = y-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the y-intercept is the y coordinate of the point (0, a) where the line crosses the y-axis.
Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15x.
What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences.
The independent variable (x) is the number of hours Svetlana tutors each session. The dependent variable (y) is the amount, in dollars, Svetlana earns for each session.
The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 (b = 15). For each session, Svetlana earns $15 for each hour she tutors.
Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x.
What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences.
The independent variable (x) is the number of hours Ethan works each visit. The dependent variable (y) is the amount, in dollars, Ethan earns for each visit.
The y-intercept is 25 (a = 25). At the start of a visit, Ethan charges a one-time fee of $25 (this is when x = 0). The slope is 20 (b = 20). For each visit, Ethan earns $20 for each hour he works.
Before we take up the discussion of linear regression and correlation, we need to examine a way to display the relation between two variables x and y. The most common and easiest way is a scatter plot. The following example illustrates a scatter plot.
In Europe and Asia, m-commerce is popular. M-commerce users have special mobile phones that work like electronic wallets as well as provide phone and Internet services. Users can do everything from paying for parking to buying a TV set or soda from a machine to banking to checking sports scores on the Internet. For the years 2000 through 2004, was there a relationship between the year and the number of m-commerce users? Construct a scatter plot. Let x = the year and let y = the number of m-commerce users, in millions.
| (year) | (# of users) |
|---|---|
| 2000 | 0.5 |
| 2002 | 20.0 |
| 2003 | 33.0 |
| 2004 | 47.0 |
Amelia plays basketball for her high school. She wants to improve to play at the college level. She notices that the number of points she scores in a game goes up in response to the number of hours she practices her jump shot each week. She records the following data:
| X (hours practicing jump shot) | Y (points scored in a game) |
|---|---|
| 5 | 15 |
| 7 | 22 |
| 9 | 28 |
| 10 | 31 |
| 11 | 33 |
| 12 | 36 |
Construct a scatter plot and state if what Amelia thinks appears to be true.
Yes, Amelia’s assumption appears to be correct. The number of points Amelia scores per game goes up when she practices her jump shot more.
A scatter plot shows the direction of a relationship between the variables. A clear direction happens when there is either:
You can determine the strength of the relationship by looking at the scatter plot and seeing how close the points are to a line, a power function, an exponential function, or to some other type of function. For a linear relationship there is an exception. Consider a scatter plot where all the points fall on a horizontal line providing a "perfect fit." The horizontal line would in fact show no relationship.
When you look at a scatterplot, you want to notice the overall pattern and any deviations from the pattern. The following scatterplot examples illustrate these concepts.
In this chapter, we are interested in scatter plots that show a linear pattern. Linear patterns are quite common. The linear relationship is strong if the points are close to a straight line, except in the case of a horizontal line where there is no relationship. If we think that the points show a linear relationship, we would like to draw a line on the scatter plot. This line can be calculated through a process called linear regression. However, we only calculate a regression line if one of the variables helps to explain or predict the other variable. If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x
Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to "fit" a straight line. This is called a Line of Best Fit or Least-Squares Line.
If you know a person's pinky (smallest) finger length, do you think you could predict that person's height? Collect data from your class (pinky finger length, in inches). The independent variable, x, is pinky finger length and the dependent variable, y, is height. For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then "by eye" draw a line that appears to "fit" the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your line so it crosses the y-axis. Using the slopes and the y-intercepts, write your equation of "best fit." Do you think everyone will have the same equation? Why or why not? According to your equation, what is the predicted height for a pinky length of 2.5 inches?
A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score?
| x (third exam score) | y (final exam score) |
|---|---|
| 65 | 175 |
| 67 | 133 |
| 71 | 185 |
| 71 | 163 |
| 66 | 126 |
| 75 | 198 |
| 67 | 153 |
| 70 | 163 |
| 71 | 159 |
| 69 | 151 |
| 69 | 159 |
SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in Table 12.4 show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet.
| X (depth in feet) | Y (maximum dive time) |
|---|---|
| 50 | 80 |
| 60 | 55 |
| 70 | 45 |
| 80 | 35 |
| 90 | 25 |
| 100 | 22 |
ŷ = 127.24 – 1.11x
At 110 feet, a diver could dive for only five minutes.
The third exam score, x, is the independent variable and the final exam score, y, is the dependent variable. We will plot a regression line that best "fits" the data. If each of you were to fit a line "by eye," you would draw different lines. We can use what is called a least-squares regression line to obtain the best fit line.
Consider the following diagram. Each point of data is of the the form (x, y) and each point ofthe line of best fit using least-squares linear regression has the form (x, ŷ).
The ŷ is read "y hat" and is the estimated value of y. It is the value of y obtained using the regression line. It is not generally equal to y from data.
The term y0 – ŷ0 = ε0 is called the "error" or residual. It is not an error in the sense of a mistake. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y. In other words, it measures the vertical distance between the actual data point and the predicted point on the line.
If the observed data point lies above the line, the residual is positive, and the line underestimates the actual data value for y. If the observed data point lies below the line, the residual is negative, and the line overestimates that actual data value for y.
In the diagram in Figure 12.11, y0 – ŷ0 = ε0 is the residual for the point shown. Here the point lies above the line and the residual is positive.
ε = the Greek letter epsilon
For each data point, you can calculate the residuals or errors, yi - ŷi = εi for i = 1, 2, 3, ..., 11.
Each |ε| is a vertical distance.
For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 ε values. If yousquare each ε and add, you get
This is called the Sum of Squared Errors (SSE).
Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation:
where and .
The sample means of the x values and the y values are and , respectively. The best fit line always passes through the point .
The slope b can be written as where sy = the standard deviation of the y values and sx = the standard deviation of the x values. r is the correlation coefficient, which is discussed in the next section.
The process of fitting the best-fit line is called linear regression. The idea behind finding the best-fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is, made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least-squares regression line .
Computer spreadsheets, statistical software, and many calculators can quickly calculate the best-fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best-fit line and create a scatterplot are shown at the end of this section.
THIRD EXAM vs FINAL EXAM EXAMPLE: The graph of the line of best fit for the third-exam/final-exam example is as follows:
The least squares regression line (best-fit line) for the third-exam/final-exam example has the equation:
Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship between the variables, then it is reasonable to use a best fit line to make predictions for y given x within the domain of x-values in the sample data, but not necessarily for x-values outside that domain. You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam. You should NOT use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x-values in the sample data, which are between 65 and 75.
The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English.
INTERPRETATION OF THE SLOPE: The slope of the best-fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average.
THIRD EXAM vs FINAL EXAM EXAMPLESlope: The slope of the line is b = 4.83.
Using the Linear Regression T Test: LinRegTTest
The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items.
Graphing the Scatterplot and Regression Line
Another way to graph the line after you create a scatter plot is to use LinRegTTest.
Besides looking at the scatter plot and seeing that a line seems reasonable, how can you tell if the line is a good predictor? Use the correlation coefficient as another indicator (besides the scatterplot) of the strength of the relationship between x and y.
The correlation coefficient, r, developed by Karl Pearson in the early 1900s, is numerical and provides a measure of strength and direction of the linear association between the independent variable x and the dependent variable y.
The correlation coefficient is calculated as
where n = the number of data points.
If you suspect a linear relationship between x and y, then r can measure how strong the linear relationship is.
What the VALUE of r tells us:
What the SIGN of r tells us
The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can quickly calculate r. The correlation coefficient r is the bottom item in the output screens for the LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions).
The variable r2 is called the coefficient of determination and is the square of the correlation coefficient, but is usually stated as a percent, rather than in decimal form. It has an interpretation in the context of the data:
Consider the third exam/final exam example introduced in the previous section
The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient r and the sample size n, together.
We perform a hypothesis test of the "significance of the correlation coefficient" to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population.
The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only have sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient.
The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is "close to zero" or "significantly different from zero". We decide this based on the sample correlation coefficient r and the sample size n.
If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is "significant."
If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is "not significant".
WHAT THE HYPOTHESES MEAN IN WORDS:
DRAWING A CONCLUSION:There are two methods of making the decision. The two methods are equivalent and give the same result.
In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05
Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.)
To calculate the p-value using LinRegTTEST:
An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR.
H0: ρ = 0
Ha: ρ ≠ 0
α = 0.05
Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.
The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of is significant or not. Compare r to the appropriate critical value in the table. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may want to use the line for prediction.
Suppose you computed r = 0.801 using n = 10 data points.df = n - 2 = 10 - 2 = 8. The critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r > positive critical value, then r issignificant. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be usedfor prediction. If you view this example on a number line, it will help you.
For a given line of best fit, you computed that r = 0.6501 using n = 12 data points and the critical value is 0.576. Can the line be used for prediction? Why or why not?
If the scatter plot looks linear then, yes, the line can be used for prediction, because r > the positive critical value.
Suppose you computed r = –0.624 with 14 data points. df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction
For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical value is 0.666. Can the line be used for prediction? Why or why not?
No, the line cannot be used for prediction, because r < the positive critical value.
Suppose you computed r = 0.776 and n = 6. df = 6 – 2 = 4. The critical values are –0.811 and 0.811. Since –0.811 < 0.776 < 0.811, r is not significant, and the line should not be used for prediction.
For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is = 0.707. Can the line be used for prediction? Why or why not?
Yes, the line can be used for prediction, because r < the negative critical value.
Consider the third exam/final exam example. The line of best fit is: ŷ = –173.51+4.83x with r = 0.6631 and there are n = 11 data points. Can the regression line be used for prediction? Given a third-exam score (x value), can we use the line to predict the final exam score (predicted y value)?
Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.
Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if r is significant and the line of best fit associated with each r can be used to predict a y value. If it helps, draw a number line.
For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not?
No, the line cannot be used for prediction no matter what the sample size is.
Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between x and y in the population.
The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this.
Recall the third exam/final exam example.
We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction.
Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75, substitute x = 73 into the equation. Then:
We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average.
Recall the third exam/final exam example.
a. What would you predict the final exam score to be for a student who scored a 66 on the third exam?
b. What would you predict the final exam score to be for a student who scored a 90 on the third exam?
b. The x values in the data are between 65 and 75. Ninety is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable.)
The process of predicting inside of the observed x values observed in the data is called interpolation. The process of predicting outside of the observed x values observed in the data is called extrapolation.
Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows:
ŷ = 72.5 + 2.8x
86.5
In some data sets, there are values (observed data points) called outliers. Outliers are observed data points that are far from the least squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point.
Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier.
Besides outliers, a sample may contain one or a few points that are called influential points. Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly.
Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them.
We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier. The standard deviation used is the standard deviation of the residuals or errors.
We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods.
In the third exam/final exam example, you can determine if there is an outlier or not. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or –1.
Graphical Identification of OutliersWith the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2s or more, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3:
As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412.
Line Y2 = –173.5 + 4.83x –2(16.4) and line Y3 = –173.5 + 4.83x + 2(16.4)
Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line.
Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers.
Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6.
The outlier appears to be at (6, 58). The expected y value on the line for the point (6, 58) is approximately 82. Fifty-eight is 24 units from 82. Twenty-four is more than two standard deviations (2s = (2)(8.6) = 17.2 ). So 82 is more than two standard deviations from 58, which makes (6, 58) a potential outlier.
In Table 12.5, the first two columns are the third-exam and final-exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x. The residuals, or errors, have been calculated in the fourth column of the table: observed y value−predicted y value = y − ŷ.
s is the standard deviation of all the y − ŷ = ε values where n = the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as:
We divide by (n – 2) because the regression model involves two estimates.
Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35 –17 16 –6 –19 9 3 –1 –10 –9 –1 .
| x | y | ŷ | y – ŷ |
|---|---|---|---|
| 65 | 175 | 140 | 175 – 140 = 35 |
| 67 | 133 | 150 | 133 – 150= –17 |
| 71 | 185 | 169 | 185 – 169 = 16 |
| 71 | 163 | 169 | 163 – 169 = –6 |
| 66 | 126 | 145 | 126 – 145 = –19 |
| 75 | 198 | 189 | 198 – 189 = 9 |
| 67 | 153 | 150 | 153 – 150 = 3 |
| 70 | 163 | 164 | 163 – 164 = –1 |
| 71 | 159 | 169 | 159 – 169 = –10 |
| 69 | 151 | 160 | 151 – 160 = –9 |
| 69 | 159 | 160 | 159 – 160 = –1 |
We are looking for all data points for which the residual is greater than 2s = 2(16.4) = 32.8 or less than –32.8. Compare these values to the residuals in column four of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35.
Numerically and graphically, we have identified the point (65, 175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on and delete the outlier, so that we can explore how it affects the results, as a learning experience.
Compute a new best-fit line and correlation coefficient using the ten remaining points: On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are:
ŷ = –355.19 + 7.39x and r = 0.9121
The new line with r = 0.9121 is a stronger correlation than the original (r = 0.6631) because r = 0.9121 is closer to one. This means that the new line is a better fit to the ten remaining data values. The line can better predict the final exam score given the third exam score.
If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following.
The squares are 352 172 162 62 192 92 32 12 102 92 12
Then, add (sum) all the |y – ŷ| squared terms using the formula
(Recall that yi – ŷi = εi.)
= 352 + 172 + 162 + 62 + 192 + 92 + 32 + 12 + 102 + 92 + 12
= 2440 = SSE. The result, SSE is the Sum of Squared Errors.
Next, calculate s, the standard deviation of all the y – ŷ = ε values where n = the total number of data points.
The calculation is .
For the third exam/final exam problem, .
Next, multiply s by 2:
If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance is at least 2s, then we would consider the data point to be "too far" from the line of best fit. We call that point a potential outlier.
For the example, if any of the |y – ŷ| values are at least 32.94, the corresponding (x, y) data point is a potential outlier.
For the third exam/final exam problem, all the |y – ŷ|'s are less than 31.29 except for the first one which is 35.
35 > 31.29 That is, |y – ŷ| ≥ (2)(s)
The point which corresponds to |y – ŷ| = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.)
When outliers are deleted, the researcher should either record that data was deleted, and why, or the researcher should provide results both with and without the deleted data. If data is erroneous and the correct values are known (e.g., student one actually scored a 70 instead of a 65), then this correction can be made to the data.
ŷ = –355.19 + 7.39x and r = 0.9121
Using this new line of best fit (based on the remaining ten data points in the third exam/final exam example), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line?
Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam.
The data points for the graph from the third exam/final exam example are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10.
ŷ = 1.04 + 2.96x; 30.64
The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. In the following table, x is the year and y is the CPI.
| x | y | x | y |
|---|---|---|---|
| 1915 | 10.1 | 1969 | 36.7 |
| 1926 | 17.7 | 1975 | 49.3 |
| 1935 | 13.7 | 1979 | 72.6 |
| 1940 | 14.7 | 1980 | 82.4 |
| 1947 | 24.1 | 1986 | 109.6 |
| 1952 | 26.5 | 1991 | 130.7 |
| 1964 | 31.0 | 1999 | 166.6 |
In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician should prefer to use other methods to fit a curve to this data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate.
If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt; our data is taken from the column entitled "Annual Avg." (third column from the right). For example you could add more current years of data. Try adding the more recent years: 2004: CPI = 188.9; 2008: CPI = 215.3; 2011: CPI = 224.9. See how it affects the model. (Check: ŷ = –4436 + 2.295x; r = 0.9018. Is r significant? Is the fit better with the addition of the new points?)
The following table shows economic development measured in per capita income PCINC.
| Year | PCINC | Year | PCINC |
|---|---|---|---|
| 1870 | 340 | 1920 | 1050 |
| 1880 | 499 | 1930 | 1170 |
| 1890 | 592 | 1940 | 1364 |
| 1900 | 757 | 1950 | 1836 |
| 1910 | 927 | 1960 | 2132 |
a. The independent variable (x) is the year and the dependent variable (y) is the per capita income.
b.
c. ŷ = 18.61x – 34574; r = 0.9732
d. At df = 8, the critical value is 0.632. The r value is significant because it is greater than the critical value.
e. There does appear to be a linear relationship between the variables.
f. The coefficient of determination is 0.947, which means that 94.7% of the variation in PCINC is explained by the variation in the years.
g. and h. The slope of the regression equation is 18.61, and it means that per capita income increases by $18.61 for each passing year. ŷ = 785 when the year is 1900, and ŷ = 2,646 when the year is 2000.
i. There do not appear to be any outliers.
| Degrees of Freedom: n – 2 | Critical Values: (+ and –) |
|---|---|
| 1 | 0.997 |
| 2 | 0.950 |
| 3 | 0.878 |
| 4 | 0.811 |
| 5 | 0.754 |
| 6 | 0.707 |
| 7 | 0.666 |
| 8 | 0.632 |
| 9 | 0.602 |
| 10 | 0.576 |
| 11 | 0.555 |
| 12 | 0.532 |
| 13 | 0.514 |
| 14 | 0.497 |
| 15 | 0.482 |
| 16 | 0.468 |
| 17 | 0.456 |
| 18 | 0.444 |
| 19 | 0.433 |
| 20 | 0.423 |
| 21 | 0.413 |
| 22 | 0.404 |
| 23 | 0.396 |
| 24 | 0.388 |
| 25 | 0.381 |
| 26 | 0.374 |
| 27 | 0.367 |
| 28 | 0.361 |
| 29 | 0.355 |
| 30 | 0.349 |
| 40 | 0.304 |
| 50 | 0.273 |
| 60 | 0.250 |
| 70 | 0.232 |
| 80 | 0.217 |
| 90 | 0.205 |
| 100 | 0.195 |
Class Time:
Names:
Collect the DataUse eight members of your class for the sample. Collect bivariate data (distance an individual lives from school, the cost of supplies for the current term).
| Distance from school | Cost of supplies this term |
|---|---|
Analyze the DataEnter your data into your calculator or computer. Write the linear equation, rounding to four decimal places.
Class Time:
Names:
Collect the Data Survey ten textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook).
| Number of pages | Cost of textbook |
|---|---|
Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places.
Class Time:
Names:
Collect the DataUse the most recent April issue of Consumer Reports. It will give the total fuel efficiency (in miles per gallon) and weight (in pounds) of new model cars with automatic transmissions. We will use this data to determine the relationship, if any, between the fuel efficiency of a car and its weight.
| Weight | Fuel Efficiency |
|---|---|
Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to 4 decimal places.
The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form y = mx + b, where m and b are constants, x is the independent variable, y is the dependent variable. In a statistical context, a linear equation is written in the form y = a + bx, where a and b are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation y = a + bx, the constant b that multiplies the x variable (b is called a coefficient) is called as the slope. The slope describes the rate of change between the independent and dependent variables; in other words, the rate of change describes the change that occurs in the dependent variable as the independent variable is changed. In the equation y = a + bx, the constant a is called as the y-intercept. Graphically, the y-intercept is the y coordinate of the point where the graph of the line crosses the y axis. At this point x = 0.
The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average. The y-intercept is used to describe the dependent variable when the independent variable equals zero. Graphically, the slope is represented by three line types in elementary statistics.
Scatter plots are particularly helpful graphs when we want to see if there is a linear relationship among data points. They indicate both the direction of the relationship between the x variables and the y variables, and the strength of the relationship. We calculate the strength of the relationship between an independent variable and a dependent variable using linear regression.
A regression line, or a line of best fit, can be drawn on a scatter plot and used to predict outcomes for the x and y variables in a given data set or sample data. There are several ways to find a regression line, but usually the least-squares regression line is used because it creates a uniform line. Residuals, also called “errors,” measure the distance from the actual value of y and the estimated value of y. The Sum of Squared Errors, when set to its minimum, calculates the points on the line of best fit. Regression lines can be used to predict values within the given set of data, but should not be used to make predictions for values outside the set of data.
The correlation coefficient r measures the strength of the linear association between x and y. The variable r has to be between –1 and +1. When r is positive, the x and y will tend to increase and decrease together. When r is negative, x will increase and y will decrease, or the opposite, x will decrease and y will increase. The coefficient of determination r2, is equal to the square of the correlation coefficient. When expressed as a percent, r2 represents the percent of variation in the dependent variable y that can be explained by variation in the independent variable x using the regression line.
Linear regression is a procedure for fitting a straight line of the form ŷ = a + bx to data. The conditions for regression are:
The slope b and intercept a of the least-squares line estimate the slope β and intercept α of the population (true) regression line. To estimate the population standard deviation of y, σ, use the standard deviation of the residuals, s. . The variable ρ (rho) is the population correlation coefficient. To test the null hypothesis H0: ρ = hypothesized value, use a linear regression t-test. The most common null hypothesis is H0: ρ = 0 which indicates there is no linear relationship between x and y in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest).
After determining the presence of a strong correlation coefficient and calculating the line of best fit, you can use the least squares regression line to make predictions about your data.
To determine if a point is an outlier, do one of the following:
where s is the standard deviation of the residuals
If any point is above y2or below y3 then the point is considered to be an outlier.
y = a + bx where a is the y-intercept and b is the slope. The variable x is the independent variable and y is the dependent variable.
Least Squares Line or Line of Best Fit:
where
a = y-intercept
b = slope
Standard deviation of the residuals:
where
SSE = sum of squared errors
n = the number of data points
Use the following information to answer the next three exercises. A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour.
What are the dependent and independent variables?
Find the equation that expresses the total fee in terms of the number of hours the equipment is rented.
Graph the equation from Exercise 12.2.
Find the equation that expresses the total fee in terms of the number of days the payment is late.
Graph the equation from Exercise 12.4.
Is the equation y = 10 + 5x – 3x2 linear? Why or why not?
Which of the following equations are linear?
a. y = 6x + 8
b. y + 7 = 3x
c. y – x = 8x2
d. 4y = 8
Does the graph show a linear equation? Why or why not?
Table 12.12 contains real data for the first two decades of AIDS reporting.
| Year | # AIDS cases diagnosed | # AIDS deaths |
| Pre-1981 | 91 | 29 |
| 1981 | 319 | 121 |
| 1982 | 1,170 | 453 |
| 1983 | 3,076 | 1,482 |
| 1984 | 6,240 | 3,466 |
| 1985 | 11,776 | 6,878 |
| 1986 | 19,032 | 11,987 |
| 1987 | 28,564 | 16,162 |
| 1988 | 35,447 | 20,868 |
| 1989 | 42,674 | 27,591 |
| 1990 | 48,634 | 31,335 |
| 1991 | 59,660 | 36,560 |
| 1992 | 78,530 | 41,055 |
| 1993 | 78,834 | 44,730 |
| 1994 | 71,874 | 49,095 |
| 1995 | 68,505 | 49,456 |
| 1996 | 59,347 | 38,510 |
| 1997 | 47,149 | 20,736 |
| 1998 | 38,393 | 19,005 |
| 1999 | 25,174 | 18,454 |
| 2000 | 25,522 | 17,347 |
| 2001 | 25,643 | 17,402 |
| 2002 | 26,464 | 16,371 |
| Total | 802,118 | 489,093 |
Use the columns "year" and "# AIDS cases diagnosed. Why is “year” the independent variable and “# AIDS cases diagnosed.” the dependent variable (instead of the reverse)?
What are the independent and dependent variables?
What is the y-intercept and what is the slope? Interpret them using complete sentences.
What are the independent and dependent variables?
How many pounds of soil does the shoreline lose in a year?
What is the y-intercept? Interpret its meaning.
What are the slope and y-intercept? Interpret their meaning.
If you owned this stock, would you want a positive or negative slope? Why?
Does the scatter plot appear linear? Strong or weak? Positive or negative?
Does the scatter plot appear linear? Strong or weak? Positive or negative?
Does the scatter plot appear linear? Strong or weak? Positive or negative?
Use the following information to answer the next five exercises. A random sample of ten professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars).
| x | y | x | y |
|---|---|---|---|
| 0 | 2 | 5 | 12 |
| 3 | 8 | 4 | 9 |
| 2 | 7 | 3 | 9 |
| 1 | 3 | 0 | 3 |
| 5 | 13 | 4 | 10 |
Draw a scatter plot of the data.
Use regression to find the equation for the line of best fit.
Draw the line of best fit on the scatter plot.
What is the slope of the line of best fit? What does it represent?
What is the y-intercept of the line of best fit? What does it represent?
What does an r value of zero mean?
When n = 2 and r = 1, are the data significant? Explain.
When n = 100 and r = -0.89, is there a significant correlation? Explain.
When testing the significance of the correlation coefficient, what is the null hypothesis?
When testing the significance of the correlation coefficient, what is the alternative hypothesis?
If the level of significance is 0.05 and the p-value is 0.04, what conclusion can you draw?
Use the following information to answer the next two exercises. An electronics retailer used regression to find a simple model to predict sales growth in the first quarter of the new year (January through March). The model is good for 90 days, where x is the day. The model can be written as follows:
ŷ = 101.32 + 2.48x where ŷ is in thousands of dollars.
What would you predict the sales to be on day 60?
What would you predict the sales to be on day 90?
ŷ = 1350 – 1.2x where x is the number of hours and ŷ represents the number of acres left to mow.
How many acres will be left to mow after 20 hours of work?
How many acres will be left to mow after 100 hours of work?
How many hours will it take to mow all of the lawns? (When is ŷ = 0?)
Table 12.14 contains real data for the first two decades of AIDS reporting.
| Year | # AIDS cases diagnosed | # AIDS deaths |
| Pre-1981 | 91 | 29 |
| 1981 | 319 | 121 |
| 1982 | 1,170 | 453 |
| 1983 | 3,076 | 1,482 |
| 1984 | 6,240 | 3,466 |
| 1985 | 11,776 | 6,878 |
| 1986 | 19,032 | 11,987 |
| 1987 | 28,564 | 16,162 |
| 1988 | 35,447 | 20,868 |
| 1989 | 42,674 | 27,591 |
| 1990 | 48,634 | 31,335 |
| 1991 | 59,660 | 36,560 |
| 1992 | 78,530 | 41,055 |
| 1993 | 78,834 | 44,730 |
| 1994 | 71,874 | 49,095 |
| 1995 | 68,505 | 49,456 |
| 1996 | 59,347 | 38,510 |
| 1997 | 47,149 | 20,736 |
| 1998 | 38,393 | 19,005 |
| 1999 | 25,174 | 18,454 |
| 2000 | 25,522 | 17,347 |
| 2001 | 25,643 | 17,402 |
| 2002 | 26,464 | 16,371 |
| Total | 802,118 | 489,093 |
Graph “year” versus “# AIDS cases diagnosed” (plot the scatter plot). Do not include pre-1981 data.
Perform linear regression. What is the linear equation? Round to the nearest whole number.
Write the equations:
Solve.
Does the line seem to fit the data? Why or why not?
What does the correlation imply about the relationship between time (years) and the number of diagnosed AIDS cases reported in the U.S.?
Plot the two given points on the following graph. Then, connect the two points to form the regression line.
Obtain the graph on your calculator or computer.
Write the equation: ŷ= ____________
Hand draw a smooth curve on the graph that shows the flow of the data.
Does the line seem to fit the data? Why or why not?
Do you think a linear fit is best? Why or why not?
What does the correlation imply about the relationship between time (years) and the number of diagnosed AIDS cases reported in the U.S.?
Graph “year” vs. “# AIDS cases diagnosed.” Do not include pre-1981. Label both axes with words. Scale both axes.
Enter your data into your calculator or computer. The pre-1981 data should not be included. Why is that so?
Write the linear equation, rounding to four decimal places:
Calculate the following:
Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69.
Do there appear to be any outliers?
A point is removed, and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why?
What effect did the potential outlier have on the line of best fit?
Are you more or less confident in the predictive ability of the new line of best fit?
The Sum of Squared Errors for a data set of 18 numbers is 49. What is the standard deviation?
The Standard Deviation for the Sum of Squared Errors for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier?
For each of the following situations, state the independent variable and the dependent variable.
Piece-rate systems are widely debated incentive payment plans. In a recent study of loan officer effectiveness, the following piece-rate system was examined:
| % of goal reached | < 80 | 80 | 100 | 120 |
| Incentive | n/a | $4,000 with an additional $125 added per percentage point from 81–99% | $6,500 with an additional $125 added per percentage point from 101–119% | $9,500 with an additional $125 added per percentage point starting at 121% |
If a loan officer makes 95% of his or her goal, write the linear function that applies based on the incentive plan table. In context, explain the y-intercept and slope.
The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. Table 12.16 shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data.
| Year | Cuba’s PPP | Year | Cuba’s PPP |
|---|---|---|---|
| 1999 | 1,700 | 2006 | 4,000 |
| 2000 | 1,700 | 2007 | 11,000 |
| 2002 | 2,300 | 2008 | 9,500 |
| 2003 | 2,900 | 2009 | 9,700 |
| 2004 | 3,000 | 2010 | 9,900 |
| 2005 | 3,500 |
The following table shows the poverty rates and cell phone usage in the United States. Construct a scatter plot of the data
| Year | Poverty Rate | Cellular Usage per Capita |
|---|---|---|
| 2003 | 12.7 | 54.67 |
| 2005 | 12.6 | 74.19 |
| 2007 | 12 | 84.86 |
| 2009 | 12 | 90.82 |
Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data.
| School | Mid-Career Salary (in thousands) | Yearly Tuition |
|---|---|---|
| Princeton | 137 | 28,540 |
| Harvey Mudd | 135 | 40,133 |
| CalTech | 127 | 39,900 |
| US Naval Academy | 122 | 0 |
| West Point | 120 | 0 |
| MIT | 118 | 42,050 |
| Lehigh University | 118 | 43,220 |
| NYU-Poly | 117 | 39,565 |
| Babson College | 117 | 40,400 |
| Stanford | 114 | 54,506 |
If the level of significance is 0.05 and the p-value is 0.06, what conclusion can you draw?
If there are 15 data points in a set of data, what is the number of degree of freedom?
What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern?
Explain what it means when a correlation has an r2 of 0.72.
Can a coefficient of determination be negative? Why or why not?
If the level of significance is 0.05 and the p-value is 0.06, what conclusion can you draw?
If there are 15 data points in a set of data, what is the number of degree of freedom?
Recently, the annual number of driver deaths per 100,000 for the selected age groups was as follows:
| Age | Number of Driver Deaths per 100,000 |
|---|---|
| 17.5 | 38 |
| 22 | 36 |
| 29.5 | 24 |
| 44.5 | 20 |
| 64.5 | 18 |
| 80 | 28 |
Table 12.20 shows the life expectancy for an individual born in the United States in certain years.
| Year of Birth | Life Expectancy |
|---|---|
| 1930 | 59.7 |
| 1940 | 62.9 |
| 1950 | 70.2 |
| 1965 | 69.7 |
| 1973 | 71.4 |
| 1982 | 74.5 |
| 1987 | 75 |
| 1992 | 75.7 |
| 2010 | 78.7 |
The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition ten, for various pages is given in Table 12.21
| Page number | Maximum value ($) |
|---|---|
| 4 | 16 |
| 14 | 19 |
| 25 | 15 |
| 32 | 17 |
| 43 | 19 |
| 57 | 15 |
| 72 | 16 |
| 85 | 15 |
| 90 | 17 |
Table 12.22 gives the gold medal times for every other Summer Olympics for the women’s 100-meter freestyle (swimming).
| Year | Time (seconds) |
|---|---|
| 1912 | 82.2 |
| 1924 | 72.4 |
| 1932 | 66.8 |
| 1952 | 66.8 |
| 1960 | 61.2 |
| 1968 | 60.0 |
| 1976 | 55.65 |
| 1984 | 55.92 |
| 1992 | 54.64 |
| 2000 | 53.8 |
| 2008 | 53.1 |
| State | # letters in name | Year entered the Union | Rank for entering the Union | Area (square miles) |
|---|---|---|---|---|
| Alabama | 7 | 1819 | 22 | 52,423 |
| Colorado | 8 | 1876 | 38 | 104,100 |
| Hawaii | 6 | 1959 | 50 | 10,932 |
| Iowa | 4 | 1846 | 29 | 56,276 |
| Maryland | 8 | 1788 | 7 | 12,407 |
| Missouri | 8 | 1821 | 24 | 69,709 |
| New Jersey | 9 | 1787 | 3 | 8,722 |
| Ohio | 4 | 1803 | 17 | 44,828 |
| South Carolina | 13 | 1788 | 8 | 32,008 |
| Utah | 4 | 1896 | 45 | 84,904 |
| Wisconsin | 9 | 1848 | 30 | 65,499 |
We are interested in whether or not the number of letters in a state name depends upon the year the state entered the Union.
The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level).
| Height (in feet) | Stories |
|---|---|
| 1,050 | 57 |
| 428 | 28 |
| 362 | 26 |
| 529 | 40 |
| 790 | 60 |
| 401 | 22 |
| 380 | 38 |
| 1,454 | 110 |
| 1,127 | 100 |
| 700 | 46 |
Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration.
Percent return:74; 66; 81; 52; 73; 62; 52; 45; 62; 46; 60; 46; 38
The following table shows data on average per capita wine consumption and heart disease rate in a random sample of 10 countries.
| Yearly wine consumption in liters | 2.5 | 3.9 | 2.9 | 2.4 | 2.9 | 0.8 | 9.1 | 2.7 | 0.8 | 0.7 |
| Death from heart diseases | 221 | 167 | 131 | 191 | 220 | 297 | 71 | 172 | 211 | 300 |
The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days:
| Swim Time | Heart Rate |
|---|---|
| 34.12 | 144 |
| 35.72 | 152 |
| 34.72 | 124 |
| 34.05 | 140 |
| 34.13 | 152 |
| 35.73 | 146 |
| 36.17 | 128 |
| 35.57 | 136 |
| 35.37 | 144 |
| 35.57 | 148 |
A researcher is investigating whether non-white minorities commit a disproportionate number of homicides. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are white males.
| White Males | Homicide rate per 100,000 people |
|---|---|
| 558,724 | 8.6 |
| 538,584 | 8.9 |
| 519,171 | 8.52 |
| 500,457 | 8.89 |
| 482,418 | 13.07 |
| 465,029 | 14.57 |
| 448,267 | 21.36 |
| 432,109 | 28.03 |
| 416,533 | 31.49 |
| 401,518 | 37.39 |
| 387,046 | 46.26 |
| 373,095 | 47.24 |
| 359,647 | 52.33 |
| School | Mid-Career Salary (in thousands) | Yearly Tuition |
|---|---|---|
| Princeton | 137 | 28,540 |
| Harvey Mudd | 135 | 40,133 |
| CalTech | 127 | 39,900 |
| US Naval Academy | 122 | 0 |
| West Point | 120 | 0 |
| MIT | 118 | 42,050 |
| Lehigh University | 118 | 43,220 |
| NYU-Poly | 117 | 39,565 |
| Babson College | 117 | 40,400 |
| Stanford | 114 | 54,506 |
Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer.
The average number of people in a family that received welfare for various years is given in Table 12.29.
| Year | Welfare family size |
|---|---|
| 1969 | 4.0 |
| 1973 | 3.6 |
| 1975 | 3.2 |
| 1979 | 3.0 |
| 1983 | 3.0 |
| 1988 | 3.0 |
| 1991 | 2.9 |
The percent of female wage and salary workers who are paid hourly rates is given in Table 12.30 for the years 1979 to 1992.
| Year | Percent of workers paid hourly rates |
|---|---|
| 1979 | 61.2 |
| 1980 | 60.7 |
| 1981 | 61.3 |
| 1982 | 61.3 |
| 1983 | 61.8 |
| 1984 | 61.7 |
| 1985 | 61.8 |
| 1986 | 62.0 |
| 1987 | 62.7 |
| 1990 | 62.8 |
| 1992 | 62.9 |
| Size (ounces) | Cost ($) | Cost per ounce |
|---|---|---|
| 16 | 3.99 | |
| 32 | 4.99 | |
| 64 | 5.99 | |
| 200 | 10.99 |
According to a flyer by a Prudential Insurance Company representative, the costs of approximate probate fees and taxes for selected net taxable estates are as follows:
| Net Taxable Estate ($) | Approximate Probate Fees and Taxes ($) |
|---|---|
| 600,000 | 30,000 |
| 750,000 | 92,500 |
| 1,000,000 | 203,000 |
| 1,500,000 | 438,000 |
| 2,000,000 | 688,000 |
| 2,500,000 | 1,037,000 |
| 3,000,000 | 1,350,000 |
The following are advertised sale prices of color televisions at Anderson’s.
| Size (inches) | Sale Price ($) |
|---|---|
| 9 | 147 |
| 20 | 197 |
| 27 | 297 |
| 31 | 447 |
| 35 | 1177 |
| 40 | 2177 |
| 60 | 2497 |
Table 12.34 shows the average heights for American boy s in 1990.
| Age (years) | Height (cm) |
|---|---|
| birth | 50.8 |
| 2 | 83.8 |
| 3 | 91.4 |
| 5 | 106.6 |
| 7 | 119.3 |
| 10 | 137.1 |
| 14 | 157.5 |
| State | # letters in name | Year entered the Union | Ranks for entering the Union | Area (square miles) |
|---|---|---|---|---|
| Alabama | 7 | 1819 | 22 | 52,423 |
| Colorado | 8 | 1876 | 38 | 104,100 |
| Hawaii | 6 | 1959 | 50 | 10,932 |
| Iowa | 4 | 1846 | 29 | 56,276 |
| Maryland | 8 | 1788 | 7 | 12,407 |
| Missouri | 8 | 1821 | 24 | 69,709 |
| New Jersey | 9 | 1787 | 3 | 8,722 |
| Ohio | 4 | 1803 | 17 | 44,828 |
| South Carolina | 13 | 1788 | 8 | 32,008 |
| Utah | 4 | 1896 | 45 | 84,904 |
| Wisconsin | 9 | 1848 | 30 | 65,499 |
We are interested in whether there is a relationship between the ranking of a state and the area of the state.
Data from the Centers for Disease Control and Prevention.
Data from the National Center for HIV, STD, and TB Prevention.
Data from the Centers for Disease Control and Prevention.
Data from the National Center for HIV, STD, and TB Prevention.
Data from the United States Census Bureau. Available online at http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html
Data from the National Center for Health Statistics.
Data from the House Ways and Means Committee, the Health and Human Services Department.
Data from Microsoft Bookshelf.
Data from the United States Department of Labor, the Bureau of Labor Statistics.
Data from the Physician’s Handbook, 1990.
Data from the United States Department of Labor, the Bureau of Labor Statistics.
dependent variable: fee amount; independent variable: time
y = 6x + 8, 4y = 8, and y + 7 = 3x are all linear equations.
The number of AIDS cases depends on the year. Therefore, year becomes the independent variable and the number of AIDS cases is the dependent variable.
The y-intercept is 50 (a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 (b = 100). For each session, the company charges $100 for each hour they clean.
12,000 pounds of soil
The slope is –1.5 (b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y-intercept is $15 (a = 15). This means the price of stock before the trading day was $15.
The data appear to be linear with a strong, positive correlation.
The data appear to have no correlation.
ŷ = 2.23 + 1.99x
The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year.
It means that there is no correlation between the data sets.
Yes, there are enough data points and the value of r is strong enough to show that there is a strong negative correlation between the data sets.
Ha: ρ ≠ 0
$250,120
1,326 acres
1,125 hours, or when x = 1,125
Check student’s solution.
Also, the correlation r = 0.4526. If r is compared to the value in the 95% Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But the scatter plot indicates otherwise.
= 3,448,225 + 1750x
There was an increase in AIDS cases diagnosed until 1993. From 1993 through 2002, the number of AIDS cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data.
Since there is no linear association between year and # of AIDS cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable “causes” the other variable.
We don’t know if the pre-1981 data was collected from a single year. So we don’t have an accurate x value for this figure.
Regression equation: ŷ (#AIDS Cases) = –3,448,225 + 1749.777 (year)
| Coefficients | |
|---|---|
| Intercept | –3,448,225 |
| X Variable 1 | 1,749.777 |
Yes, there appears to be an outlier at (6, 58).
The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate is a predictor for the data.
s = 1.75
Check student’s solution.
For graph: check student’s solution. Note that tuition is the independent variable and salary is the dependent variable.
13
It means that 72% of the variation in the dependent variable (y) can be explained by the variation in the independent variable (x).
We do not reject the null hypothesis. There is not sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero.
| Age | Number of Driver Deaths per 100,000 |
|---|---|
| 16–19 | 38 |
| 20–24 | 36 |
| 25–34 | 24 |
| 35–54 | 20 |
| 55–74 | 18 |
| 75+ | 28 |
As the page number increases by one page, the discount decreases by $0.01412
a. and b. Check student’s solution.
c. The slope of the regression line is -0.3179 with a y-intercept of 32.966. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 31% new sparrow hawks, which doesn’t make sense since if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 0.3179%.
d. If we examine r2, we see that only 50.238% of the variation in the percent of new birds is explained by the model and the correlation coefficient, r = 0.71 only indicates a somewhat strong correlation between returning and new percentages.
e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66%, our observed new percentage, 6%, is almost 6% less than the predicted new value of 11.98%. If we remove this data pair, we see only an adjusted slope of -0.2723 and an adjusted intercept of 30.606. In other words, even though this data generates the largest residual, it is not an outlier, nor is the data pair an influential point.
f. If there are 70% returning birds, we would expect to see y = -0.2723(70) + 30.606 = 0.115 or 11.5% new birds in the colony.
If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160 with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set.
| Size (ounces) | Cost ($) | cents/oz |
|---|---|---|
| 16 | 3.99 | 24.94 |
| 32 | 4.99 | 15.59 |
| 64 | 5.99 | 9.36 |
| 200 | 10.99 | 5.50 |
| Alabama | 7 | 1819 | 22 | 52,423 |
|---|---|---|---|---|
| Colorado | 8 | 1876 | 38 | 104,100 |
| Alaska | 6 | 1959 | 51 | 656,424 |
| Iowa | 4 | 1846 | 29 | 56,276 |
| Maryland | 8 | 1788 | 7 | 12,407 |
| Missouri | 8 | 1821 | 24 | 69,709 |
| New Jersey | 9 | 1787 | 3 | 8,722 |
| Ohio | 4 | 1803 | 17 | 44,828 |
| South Carolina | 13 | 1788 | 8 | 32,008 |
| Utah | 4 | 1896 | 45 | 84,904 |
| Wisconsin | 9 | 1848 | 30 | 65,499 |
By the end of this chapter, the student should be able to:
Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her upbringing. A consumer looking for a new car might compare the average gas mileage of several models.
For hypothesis tests comparing averages between more than two groups, statisticians have developed a method called "Analysis of Variance" (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or one-way ANOVA. You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is just a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer.
The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a one-way ANOVA test, there are five basic assumptions to be fulfilled:
The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups:
H0: μ1 = μ2 = μ3 = ... = μk
Ha: At least two of the group means μ1, μ2, μ3, ..., μk are not equal.
The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (red box plots), H0: μ1 = μ2 = μ3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations.
If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots).
The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.
For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F4,10.
The F distribution is derived from the Student's t-distribution. The values of the F distribution are squares of the corresponding values of the t-distribution. One-Way ANOVA expands the t-test for comparing more than two groups. The scope of that derivation is beyond the level of this course.
To calculate the F ratio, two estimates of the variance are made.
To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics.
MS means "mean square." MSbetween is the variance between groups, and MSwithin is the variance within groups.
Calculation of Sum of Squares and Mean Square
MSbetween and MSwithin can be written as follows:
The one-way ANOVA test depends on the fact that MSbetween can be influenced by population differences among means of the several groups. Since MSwithin compares values of each group to its own group mean, the fact that group means might be different does not affect MSwithin.
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MSbetween and MSwithin should both estimate the same value.
The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances.
F-Ratio or F Statistic
If MSbetween and MSwithin estimate the same value (following the belief that H0 is true), then the F-ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MSbetween consists of the population variance plus a variance produced from the differences between the samples. MSwithin is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin.Then the F-ratio will be larger than one. However, if the population effect is small, it is not unlikely that MSwithin will be larger in a given sample.
The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F-ratio can be written as:
F-Ratio Formula when the groups are the same size
Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software.
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor |
SS(Factor) | k – 1 | MS(Factor) = SS(Factor)/(k – 1) | F = MS(Factor)/MS(Error) |
| Error |
SS(Error) | n – k | MS(Error) = SS(Error)/(n – k) | |
| Total | SS(Total) | n – 1 |
Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table 13.2.
| Plan 1: n1 = 4 | Plan 2: n2 = 3 | Plan 3: n3 = 3 |
|---|---|---|
| 5 | 3.5 | 8 |
| 4.5 | 7 | 4 |
| 4 | 3.5 | |
| 3 | 4.5 |
s1 = 16.5, s2 =15, s3 = 15.7
Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test.
where n1 = 4, n2 = 3, n3 = 3 and n = n1 + n2 + n3 = 10
One-Way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between) and SS(Error) = SS(Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively).
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor |
SS(Factor) |
k – 1 |
MS(Factor) |
F = |
| Error |
SS(Error) |
n – k |
MS(Error) |
|
| Total | SS(Total) |
n – 1 |
As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments
All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants:
| Bare: n1 = 3 | Ground Cover: n2 = 3 | Plastic: n3 = 3 | Straw: n4 = 3 | Compost: n5 = 3 |
|---|---|---|---|---|
| 2,625 | 5,348 | 6,583 | 7,285 | 6,277 |
| 2,997 | 5,682 | 8,560 | 6,897 | 7,818 |
| 4,915 | 5,482 | 3,830 | 9,230 | 8,677 |
Enter the data into lists L1, L2, L3, L4 and L5. Press STAT and arrow over to TESTS. Arrow down to ANOVA. Press ENTER and enter L1, L2, L3, L4, L5). Press ENTER. The table was filled in with the results from the calculator.
One-Way ANOVA table:
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 36,648,561 | 5 – 1 = 4 | ||
| Error (Within) | 20,446,726 | 15 – 5 = 10 | ||
| Total | 57,095,287 | 15 – 1 = 14 |
The one-way ANOVA hypothesis test is always right-tailed because larger F-values are way out in the right tail of the F-distribution curve and tend to make us reject H0.
The notation for the F distribution is F ~ Fdf(num),df(denom)
where df(num) = dfbetween and df(denom) = dfwithin
The mean for the F distribution is
Here are some facts about the F distribution.
Let’s return to the slicing tomato exercise in The F Distribution and the F-Ratio. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
The null and alternative hypotheses are:
H0: μ1 = μ2 = μ3 = μ4 = μ5
Ha: μi ≠ μj some i ≠ j
The one-way ANOVA results are shown in Table 13.6
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 36,648,561 | 5 – 1 = 4 | ||
| Error (Within) | 20,446,726 | 15 – 5 = 10 | ||
| Total | 57,095,287 | 15 – 1 = 14 |
Distribution for the test: F4,10
df(num) = 5 – 1 = 4
df(denom) = 15 – 5 = 10
Test statistic: F = 4.4810
Probability Statement: p-value = P(F > 4.481) = 0.0248.
Compare α and the p-value: α = 0.05, p-value = 0.0248
Make a decision: Since α > p-value, we reject H0.
Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields.
To find these results on the calculator:
Press STAT. Press 1:EDIT. Put the data into the lists L1, L2, L3, L4, L5.
Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter L1, L2, L3, L4, L5). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p-value of the test.
The calculator displays:
MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table 13.7 shows various colony counts from different patients who may or may not have MRSA.
| Conc = 0.6 | Conc = 0.8 | Conc = 1.0 | Conc = 1.2 | Conc = 1.4 |
|---|---|---|---|---|
| 9 | 16 | 22 | 30 | 27 |
| 66 | 93 | 147 | 199 | 168 |
| 98 | 82 | 120 | 148 | 132 |
Plot of the data for the different concentrations:
Test whether the mean number of colonies are the same or are different. Construct the ANOVA table (by hand or by using a TI-83, 83+, or 84+ calculator), find the p-value, and state your conclusion. Use a 5% significance level.
While there are differences in the spreads between the groups (see Figure 13.6), the differences do not appear to be big enough to cause concern.
We test for the equality of mean number of colonies:
H0 : μ1 = μ2 = μ3 = μ4 = μ5
Ha: μi ≠ μj some i ≠ j
The one-way ANOVA table results are shown in Table 13.8.
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 10,233 | 5 – 1 = 4 | ||
| Error (Within) | 41,949 | 15 – 5 = 10 | ||
| Total | 52,182 | 15 – 1 = 14 |
Distribution for the test: F4,10
Probability Statement: p-value = P(F > 0.6099) = 0.6649.
Compare α and the p-value: α = 0.05, p-value = 0.669, α > p-value
Make a decision: Since α > p-value, we do not reject H0.
Conclusion: At the 5% significance level, there is insufficient evidence from these data that different levels of tryptone will cause a significant difference in the mean number of bacterial colonies formed.
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.9.
| Sorority 1 | Sorority 2 | Sorority 3 | Sorority 4 |
|---|---|---|---|
| 2.17 | 2.63 | 2.63 | 3.79 |
| 1.85 | 1.77 | 3.78 | 3.45 |
| 2.83 | 3.25 | 4.00 | 3.08 |
| 1.69 | 1.86 | 2.55 | 2.26 |
| 3.33 | 2.21 | 2.45 | 3.18 |
Using a significance level of 1%, is there a difference in mean grades among the sororities?
Let μ1, μ2, μ3, μ4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.
This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations.
H0: μ1 = μ2 = μ3 = μ4
Ha: Not all of the means μ1, μ2, μ3, μ4 are equal.
Distribution for the test: F3,16
where k = 4 groups and n = 20 samples in total
df(num)= k – 1 = 4 – 1 = 3
df(denom) = n – k = 20 – 4 = 16
Calculate the test statistic: F = 2.23
Graph:
Probability statement: p-value = P(F > 2.23) = 0.1241
Compare α and the p-value: α = 0.01
Make a decision: Since α < p-value, you cannot reject H0.
Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.
Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4).
The calculator displays the F statistic, the p-value and the values for the one-way ANOVA table:
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.10.
| Basketball | Baseball | Hockey | Lacrosse |
|---|---|---|---|
| 3.6 | 2.1 | 4.0 | 2.0 |
| 2.9 | 2.6 | 2.0 | 3.6 |
| 2.5 | 3.9 | 2.6 | 3.9 |
| 3.3 | 3.1 | 3.2 | 2.7 |
| 3.8 | 3.4 | 3.2 | 2.5 |
Use a significance level of 5%, and determine if there is a difference in GPA among the teams.
With a p-value of 0.9271, we decline to reject the null hypothesis. There is not sufficient evidence to conclude that there is a difference among the GPAs for the sports teams.
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in Table 13.11.
| Tommy's Plants | Tara's Plants | Nick's Plants |
|---|---|---|
| 24 | 25 | 23 |
| 21 | 31 | 27 |
| 23 | 23 | 22 |
| 30 | 20 | 30 |
| 23 | 28 | 20 |
Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.
This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = .
First, calculate the sample mean and sample variance of each group.
| Tommy's Plants | Tara's Plants | Nick's Plants | |
|---|---|---|---|
| Sample Mean | 24.2 | 25.4 | 24.4 |
| Sample Variance | 11.7 | 18.3 | 16.3 |
Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 =
Then MSbetween = = (5)(0.413) where n = 5 is the sample size (number of plants each child grew).
Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s2 pooled
Then MSwithin = s2pooled = 15.433.
The F statistic (or F ratio) is
The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2.
The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12
The distribution for the test is F2,12 and the F statistic is F = 0.134
The p-value is P(F > 0.134) = 0.8759.
Decision: Since α = 0.03 and the p-value = 0.8759, do not reject H0. (Why?)
Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.
To calculate the p-value:
*Press 2nd DISTR
*Arrow down to Fcdf(and press ENTER.
*Enter 0.134, E99, 2, 12)
*Press ENTER
The p-value is 0.8759.
Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.3.
From the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.
From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets in Table 13.45.
Another of the uses of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers.
In order to perform a F test of two variances, it is important that the following are true:
Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here.
Suppose we sample randomly from two independent normal populations. Let and be the population variances and and be the sample variances. Let the sample sizes be n1 and n2. Since we are interested in comparing the two sample variances, we use the F ratio:
F has the distribution F ~ F(n1 – 1, n2 – 1)
where n1 – 1 are the degrees of freedom for the numerator and n2 – 1 are the degrees of freedom for the denominator.
If the null hypothesis is , then the F Ratio becomes .
The F ratio could also be . It depends on Ha and on which sample variance is larger.
If the two populations have equal variances, then and are close in value and is close to one. But if the two population variances are very different, and tend to be very different, too. Choosing as the larger sample variance causes the ratio to be greater than one. If and are far apart, then is a large number.
Therefore, if F is close to one, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than one, then the evidence is against the null hypothesis. A test of two variances may be left, right, or two-tailed.
Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor's grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%.
Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.
n1 = n2 = 30.
H0: and Ha:
Calculate the test statistic: By the null hypothesis , the F statistic is:
Distribution for the test: F29,29 where n1 – 1 = 29 and n2 – 1 = 29.
Graph: This test is left tailed.
Draw the graph labeling and shading appropriately.
Probability statement: p-value = P(F < 0.5818) = 0.0753
Compare α and the p-value: α = 0.10 α > p-value.
Make a decision: Since α > p-value, reject H0.
Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller.
Press STAT and arrow over to TESTS. Arrow down to D:2-SampFTest. Press ENTER. Arrow to Stats and press ENTER. For Sx1, n1, Sx2, and n2, enter , 30, , and 30. Press ENTER after each. Arrow to σ1: and σ2. Press ENTER. Arrow down to Calculate and press ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try Draw instead of Calculate.
The New York Choral Society divides male singers up into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different?
| Tenor1 | Bass2 | Tenor 1 | Bass 2 | Tenor 1 | Bass 2 |
|---|---|---|---|---|---|
| 69 | 72 | 67 | 72 | 68 | 67 |
| 72 | 75 | 70 | 74 | 67 | 70 |
| 71 | 67 | 65 | 70 | 64 | 70 |
| 66 | 75 | 72 | 66 | 69 | |
| 76 | 74 | 70 | 68 | 72 | |
| 74 | 72 | 68 | 75 | 71 | |
| 71 | 72 | 64 | 68 | 74 | |
| 66 | 74 | 73 | 70 | 75 | |
| 68 | 72 | 66 | 72 |
The histograms are not as normal as one might like. Plot them to verify. However, we proceed with the test in any case.
Subscripts: T1= tenor1 and B2 = bass 2
The standard deviations of the samples are sT1 = 3.3302 and sB2 = 2.7208.
The hypotheses are
and (two tailed test)
The F statistic is 1.4894 with 20 and 25 degrees of freedom.
The p-value is 0.3430. If we assume alpha is 0.05, then we cannot reject the null hypothesis.
We have no good evidence from the data that the heights of Tenor1 and Bass2 singers have different variances (despite there being a significant difference in mean heights of about 2.5 inches.)
Class Time:
Names:
| Fruits | Vegetables | Breads |
|---|---|---|
The test statistic for analysis of variance is the F-ratio.
The test statistic for analysis of variance is the F-ratio.
Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom.
Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table.
The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p-value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p-value) to the right of the statistic under the F curve.
When the data have unequal group sizes (unbalanced data), then techniques from Figure 13.2 need to be used for hand calculations. In the case of balanced data (the groups are the same size) however, simplified calculations based on group means and variances may be used. In practice, of course, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look of your data!
The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom.
dfbetween = df(num) = k – 1
dfwithin = df(denom) = n – k
MSbetween =
MSwithin =
F =
F ratio when the groups are the same size: F =
Mean of the F distribution: µ =
where:
F has the distribution F ~ F(n1 – 1, n2 – 1)
F =
If σ1 = σ2, then F =
Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they?
Write one assumption.
Write another assumption.
Write a third assumption.
Write a fourth assumption.
Write the final assumption.
State the null hypothesis for a one-way ANOVA test if there are four groups.
State the alternative hypothesis for a one-way ANOVA test if there are three groups.
When do you use an ANOVA test?
Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The one-way ANOVA results are shown in Table 13.15.
| Group 1 | Group 2 | Group 3 |
|---|---|---|
| 216 | 202 | 170 |
| 198 | 213 | 165 |
| 240 | 284 | 182 |
| 187 | 228 | 197 |
| 176 | 210 | 201 |
What is the Sum of Squares Factor?
What is the Sum of Squares Error?
What is the df for the numerator?
What is the df for the denominator?
What is the Mean Square Factor?
What is the Mean Square Error?
What is the F statistic?
| Team 1 | Team 2 | Team 3 | Team 4 |
|---|---|---|---|
| 1 | 2 | 0 | 3 |
| 2 | 3 | 1 | 4 |
| 0 | 2 | 1 | 4 |
| 3 | 4 | 0 | 3 |
| 2 | 4 | 0 | 2 |
What is SSbetween?
What is the df for the numerator?
What is MSbetween?
What is SSwithin?
What is the df for the denominator?
What is MSwithin?
What is the F statistic?
Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis?
An F statistic can have what values?
What happens to the curves as the degrees of freedom for the numerator and the denominator get larger?
Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in Table 13.17.
| Team 1 | Team 2 | Team 3 | Team 4 | Team 5 |
|---|---|---|---|---|
| 36 | 32 | 48 | 38 | 41 |
| 42 | 35 | 50 | 44 | 39 |
| 51 | 38 | 39 | 46 | 40 |
What is the df(num)?
What is the df(denom)?
What are the Sum of Squares and Mean Squares Factors?
What are the Sum of Squares and Mean Squares Errors?
What is the F statistic?
What is the p-value?
At the 5% significance level, is there a difference in the mean jump heights among the teams?
| Group A | Group B | Group C |
|---|---|---|
| 101 | 151 | 101 |
| 108 | 149 | 109 |
| 98 | 160 | 198 |
| 107 | 112 | 186 |
| 111 | 126 | 160 |
What is the df(num)?
What is the df(denom)?
What are the SSbetween and MSbetween?
What are the SSwithin and MSwithin?
What is the F Statistic?
What is the p-value?
At the 10% significance level, are the scores among the different groups different?
| Northeast | South | West | Central | East | |
|---|---|---|---|---|---|
| 16.3 | 16.9 | 16.4 | 16.2 | 17.1 | |
| 16.1 | 16.5 | 16.5 | 16.6 | 17.2 | |
| 16.4 | 16.4 | 16.6 | 16.5 | 16.6 | |
| 16.5 | 16.2 | 16.1 | 16.4 | 16.8 | |
| ________ | ________ | ________ | ________ | ________ | |
| ________ | ________ | ________ | ________ | ________ |
Enter the data into your calculator or computer.
p-value = ______
State the decisions and conclusions (in complete sentences) for the following preconceived levels of α.
α = 0.05
a. Decision: ____________________________
b. Conclusion: ____________________________
α = 0.01
a. Decision: ____________________________
b. Conclusion: ____________________________
Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an F test of two variances.
Name one assumption that must be true.
What is the other assumption that must be true?
State the null and alternative hypotheses.
What is s1 in this problem?
What is s2 in this problem?
What is n?
What is the F statistic?
What is the p-value?
Is the claim accurate?
State the null and alternative hypotheses.
What is the F Statistic?
What is the p-value?
At the 5% significance level, do we reject the null hypothesis?
State the null and alternative hypotheses.
What is the F Statistic?
At the 5% significance level, what can we say about the cyclists’ variances?
Three different traffic routes are tested for mean driving time. The entries in the table are the driving times in minutes on the three different routes. The one-way ANOVA results are shown in Table 13.20.
| Route 1 | Route 2 | Route 3 |
|---|---|---|
| 30 | 27 | 16 |
| 32 | 29 | 41 |
| 27 | 28 | 22 |
| 35 | 36 | 31 |
State SSbetween, SSwithin, and the F statistic.
Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
| Northeast | South | West | Central | East | |
|---|---|---|---|---|---|
| 16.3 | 16.9 | 16.4 | 16.2 | 17.1 | |
| 16.1 | 16.5 | 16.5 | 16.6 | 17.2 | |
| 16.4 | 16.4 | 16.6 | 16.5 | 16.6 | |
| 16.5 | 16.2 | 16.1 | 16.4 | 16.8 | |
| ________ | ________ | ________ | ________ | ________ | |
| ________ | ________ | ________ | ________ | ________ |
State the hypotheses.
H0: ____________
Ha: ____________
Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
| Northeast | South | West | Central | East | |
|---|---|---|---|---|---|
| 16.3 | 16.9 | 16.4 | 16.2 | 17.1 | |
| 16.1 | 16.5 | 16.5 | 16.6 | 17.2 | |
| 16.4 | 16.4 | 16.6 | 16.5 | 16.6 | |
| 16.5 | 16.2 | 16.1 | 16.4 | 16.8 | |
| ________ | ________ | ________ | ________ | ________ | |
| ________ | ________ | ________ | ________ | ________ |
H0: µ1 = µ2 = µ3 = µ4 = µ5
Hα: At least any two of the group means µ1, µ2, …, µ5 are not equal.
degrees of freedom – numerator: df(num) = _________
degrees of freedom – denominator: df(denom) = ________
F statistic = ________
Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in Table 13.45.
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain.
| Linda's rats | Tuan's rats | Javier's rats |
|---|---|---|
| 43.5 | 47.0 | 51.2 |
| 39.4 | 40.5 | 40.9 |
| 41.3 | 38.9 | 37.9 |
| 46.0 | 46.3 | 45.0 |
| 38.2 | 44.2 | 48.6 |
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in Table 13.24. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same.
| working-class | professional (middle incomes) | professional (wealthy) |
|---|---|---|
| 17.8 | 16.5 | 8.5 |
| 26.7 | 17.4 | 6.3 |
| 49.4 | 22.0 | 4.6 |
| 9.4 | 7.4 | 12.6 |
| 65.4 | 9.4 | 11.0 |
| 47.1 | 2.1 | 28.6 |
| 19.5 | 6.4 | 15.4 |
| 51.2 | 13.9 | 9.3 |
Examine the seven practice laps from Table 13.45. Determine whether the mean lap time is statistically the same for the seven practice laps, or if there is at least one lap that has a different mean time from the others.
Use the following information to answer the next two exercises. Table 13.25 lists the number of pages in four different types of magazines.
| home decorating | news | health | computer |
|---|---|---|---|
| 172 | 87 | 82 | 104 |
| 286 | 94 | 153 | 136 |
| 163 | 123 | 87 | 98 |
| 205 | 106 | 103 | 207 |
| 197 | 101 | 96 | 146 |
Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length.
Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same?
A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that Table 13.26 shows the results of a study.
| CNN | FOX | Local |
|---|---|---|
| 45 | 15 | 72 |
| 12 | 43 | 37 |
| 18 | 68 | 56 |
| 38 | 50 | 60 |
| 23 | 31 | 51 |
| 35 | 22 |
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Are the means for the final exams the same for all statistics class delivery types? Table 13.27 shows the scores on final exams from several randomly selected classes that used the different delivery types.
| Online | Hybrid | Face-to-Face |
|---|---|---|
| 72 | 83 | 80 |
| 84 | 73 | 78 |
| 77 | 84 | 84 |
| 80 | 81 | 81 |
| 81 | 86 | |
| 79 | ||
| 82 |
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics and Asians? Suppose that Table 13.28 shows the results of a study.
| White | Black | Hispanic | Asian |
|---|---|---|---|
| 6 | 4 | 7 | 8 |
| 8 | 1 | 3 | 3 |
| 2 | 5 | 5 | 5 |
| 4 | 2 | 4 | 1 |
| 6 | 6 | 7 |
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that Table 13.29 shows the results of a study.
| Powder | Machine Made | Hard Packed |
|---|---|---|
| 1,210 | 2,107 | 2,846 |
| 1,080 | 1,149 | 1,638 |
| 1,537 | 862 | 2,019 |
| 941 | 1,870 | 1,178 |
| 1,528 | 2,233 | |
| 1,382 |
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew.
| Paper Type/Trial | Trial 1 | Trial 2 | Trial 3 | Trial 4 |
|---|---|---|---|---|
| Heavy | 5.1 meters | 3.1 meters | 4.7 meters | 5.3 meters |
| Medium | 4 meters | 3.5 meters | 4.5 meters | 6.1 meters |
| Light | 3.1 meters | 3.3 meters | 2.1 meters | 1.9 meters |
DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds.
An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data:
| RS | SS | NS | RS | SS | NS |
|---|---|---|---|---|---|
| 12.8 | 38.4 | 35.4 | 22.4 | 23.1 | 22.6 |
| 21.6 | 32.9 | 27.4 | 27.5 | 29.4 | 40.4 |
| 14.8 | 48.5 | 19.3 | 20.3 | 16 | 34.4 |
| 23.1 | 20.9 | 41.8 | 38.7 | 20.1 | 30.4 |
| 34.6 | 11.6 | 20.3 | 26.4 | 23.3 | 14.9 |
| 19.7 | 22.3 | 37.6 | 23.7 | 22.9 | 51.8 |
| 22.6 | 30.2 | 36.9 | 26.1 | 22.5 | 33.8 |
| 29.6 | 33.4 | 37.3 | 29.5 | 15.1 | 37.9 |
| 16.4 | 26.7 | 28.2 | 38.6 | 31 | 29.5 |
| 20.3 | 39 | 23.4 | 44.4 | 16.9 | 42.4 |
| 29.3 | 12.8 | 33.7 | 23.2 | 16.1 | 36.6 |
| 14.9 | 14.6 | 29.2 | 23.6 | 10.8 | 47.4 |
| 27.3 | 12.2 | 41.7 |
The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other.
Here is a chart of the three groups:
The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms.
Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different?
Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals.
| FL | FH | ML | MH | FL | FH | ML | MH |
|---|---|---|---|---|---|---|---|
| 96.4 | 96.8 | 96.3 | 96.9 | 98.4 | 98.6 | 98.1 | 98.6 |
| 96.7 | 97.7 | 96.7 | 97 | 98.7 | 98.6 | 98.1 | 98.6 |
| 97.2 | 97.8 | 97.1 | 97.1 | 98.7 | 98.6 | 98.2 | 98.7 |
| 97.2 | 97.9 | 97.2 | 97.1 | 98.7 | 98.7 | 98.2 | 98.8 |
| 97.4 | 98 | 97.3 | 97.4 | 98.7 | 98.7 | 98.2 | 98.8 |
| 97.6 | 98 | 97.4 | 97.5 | 98.8 | 98.8 | 98.2 | 98.8 |
| 97.7 | 98 | 97.4 | 97.6 | 98.8 | 98.8 | 98.3 | 98.9 |
| 97.8 | 98 | 97.4 | 97.7 | 98.8 | 98.8 | 98.4 | 99 |
| 97.8 | 98.1 | 97.5 | 97.8 | 98.8 | 98.9 | 98.4 | 99 |
| 97.9 | 98.3 | 97.6 | 97.9 | 99.2 | 99 | 98.5 | 99 |
| 97.9 | 98.3 | 97.6 | 98 | 99.3 | 99 | 98.5 | 99.2 |
| 98 | 98.3 | 97.8 | 98 | 99.1 | 98.6 | 99.5 | |
| 98.2 | 98.4 | 97.8 | 98 | 99.1 | 98.6 | ||
| 98.2 | 98.4 | 97.8 | 98.3 | 99.2 | 98.7 | ||
| 98.2 | 98.4 | 97.9 | 98.4 | 99.4 | 99.1 | ||
| 98.2 | 98.4 | 98 | 98.4 | 99.9 | 99.3 | ||
| 98.2 | 98.5 | 98 | 98.6 | 100 | 99.4 | ||
| 98.2 | 98.6 | 98 | 98.6 | 100.8 |
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded.
| Linda's rats | Tuan's rats | Javier's rats |
|---|---|---|
| 43.5 | 47.0 | 51.2 |
| 39.4 | 40.5 | 40.9 |
| 41.3 | 38.9 | 37.9 |
| 46.0 | 46.3 | 45.0 |
| 38.2 | 44.2 | 48.6 |
Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%.
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows.
| working-class | professional (middle incomes) | professional (wealthy) |
|---|---|---|
| 17.8 | 16.5 | 8.5 |
| 26.7 | 17.4 | 6.3 |
| 49.4 | 22.0 | 4.6 |
| 9.4 | 7.4 | 12.6 |
| 65.4 | 9.4 | 11.0 |
| 47.1 | 2.1 | 28.6 |
| 19.5 | 6.4 | 15.4 |
| 51.2 | 13.9 | 9.3 |
Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups. Use a 5% significance level.
Refer to the data from Table 13.45.
Examine practice laps 3 and 4. Determine whether or not the variance in lap time is statistically the same for those practice laps.
Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines.
| home decorating | news | health | computer |
|---|---|---|---|
| 172 | 87 | 82 | 104 |
| 286 | 94 | 153 | 136 |
| 163 | 123 | 87 | 98 |
| 205 | 106 | 103 | 207 |
| 197 | 101 | 96 | 146 |
Which two magazine types do you think have the same variance in length?
Which two magazine types do you think have different variances in length?
Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that the Table 13.36 shows the results of a study.
| Saturday | Sunday | Saturday | Sunday |
|---|---|---|---|
| 75 | 44 | 62 | 137 |
| 18 | 58 | 0 | 82 |
| 150 | 61 | 124 | 39 |
| 94 | 19 | 50 | 127 |
| 62 | 99 | 31 | 141 |
| 73 | 60 | 118 | 73 |
| 89 |
Are the variances for incomes on the East Coast and the West Coast the same? Suppose that Table 13.37 shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05.
| East | West |
|---|---|
| 38 | 71 |
| 47 | 126 |
| 30 | 42 |
| 82 | 51 |
| 75 | 44 |
| 52 | 90 |
| 115 | 88 |
| 67 |
Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100 mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how?
Here are the data:
| 0 mg | 100 mg | 200 mg | 0 mg | 100 mg | 200 mg |
|---|---|---|---|---|---|
| 242 | 248 | 246 | 245 | 246 | 248 |
| 244 | 245 | 250 | 248 | 247 | 252 |
| 247 | 248 | 248 | 248 | 250 | 250 |
| 242 | 247 | 246 | 244 | 246 | 248 |
| 246 | 243 | 245 | 242 | 244 | 250 |
King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul) during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but declined significantly afterwards. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from a fourth. These spanned most of his reign. We have data on the silver content of the coins:
| First Coinage | Second Coinage | Third Coinage | Fourth Coinage |
|---|---|---|---|
| 5.9 | 6.9 | 4.9 | 5.3 |
| 6.8 | 9.0 | 5.5 | 5.6 |
| 6.4 | 6.6 | 4.6 | 5.5 |
| 7.0 | 8.1 | 4.5 | 5.1 |
| 6.6 | 9.3 | 6.2 | |
| 7.7 | 9.2 | 5.8 | |
| 7.2 | 8.6 | 5.8 | |
| 6.9 | |||
| 6.2 |
Did the silver content of the coins change over the course of Manuel’s reign?
Here are the means and variances of each coinage. The data are unbalanced.
| First | Second | Third | Fourth | |
|---|---|---|---|---|
| Mean | 6.7444 | 8.2429 | 4.875 | 5.6143 |
| Variance | 0.2953 | 1.2095 | 0.2025 | 0.1314 |
The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in 2012 Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in the 2012 season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same or not. Note that the data are not balanced, as two divisions had five teams, while one had only four.
| Division | Team | Wins |
|---|---|---|
| East | NY Yankees | 95 |
| East | Baltimore | 93 |
| East | Tampa Bay | 90 |
| East | Toronto | 73 |
| East | Boston | 69 |
| Division | Team | Wins |
|---|---|---|
| Central | Detroit | 88 |
| Central | Chicago Sox | 85 |
| Central | Kansas City | 72 |
| Central | Cleveland | 68 |
| Central | Minnesota | 66 |
| Division | Team | Wins |
|---|---|---|
| West | Oakland | 94 |
| West | Texas | 93 |
| West | LA Angels | 89 |
| West | Seattle | 75 |
Tomato Data, Marist College School of Science (unpublished student research)
Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA.
Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall, 1994.
Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 50.
Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118.
“MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012.
Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical Association, 268, 1578-1580.
“MLB Vs. Division Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/order/true.
Each population from which a sample is taken is assumed to be normal.
The populations are assumed to have equal standard deviations (or variances).
The response is a numerical value.
Ha: At least two of the group means μ1, μ2, μ3 are not equal.
4,939.2
2
2,469.6
3.7416
3
13.2
0.825
Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p-value, so it is likely that we will reject the null hypothesis.
The curves approximate the normal distribution.
ten
SS = 237.33; MS = 23.73
0.1614
two
3.6101
Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level.
The populations from which the two samples are drawn are normally distributed.
4.11
0.7159
No, at the 10% level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker.
2.8674
Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student.
0.7414
df(denom) = 15
The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups.
H0: μ1 = μ2 = μ3;
Ha: μi ≠ μj some i ≠ j.
Define μ1, μ2, μ3, as the population mean number of eggs laid by the three groups of fruit flies.
F statistic = 8.6657;
p-value = 0.0004
Decision: Since the p-value is less than the level of significance of 0.01, we reject the null hypothesis.
Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different.
Interestingly, if you perform a two sample t-test to compare the RS and NS groups they are significantly different (p = 0.0013). Similarly, SS and NS are significantly different (p = 0.0006). However, the two selected groups, RS and SS are not significantly different (p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity.
The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances.
Here is a strip chart of the silver content of the coins:
While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table:
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 37.748 | 4 – 1 = 3 | 12.5825 | 26.272 |
| Error (Within) | 11.015 | 27 – 4 = 23 | 0.4789 | |
| Total | 48.763 | 27 – 1 = 26 |
P(F > 26.272) = 0;
Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth.
Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season.
While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust.
Here is the ANOVA table for the data:
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 344.16 | 3 – 1 = 2 | 172.08 | 26.272 |
| Error (Within) | 1,219.55 | 14 – 3 = 11 | 110.87 | 1.5521 |
| Total | 1,563.71 | 14 – 1 = 13 |
P(F > 1.5521) = 0.2548
These review exercises are designed to provide extra practice on concepts learned before a particular chapter. For example, the review exercises for Chapter 3, cover material learned in chapters 1 and 2.
Use the following information to answer the next six exercises: In a survey of 100 stocks on NASDAQ, the average percent increase for the past year was 9% for NASDAQ stocks.
1. The “average increase” for all NASDAQ stocks is the:
2. All of the NASDAQ stocks are the:
3. Nine percent is the:
4. The 100 NASDAQ stocks in the survey are the:
5. The percent increase for one stock in the survey is the:
6. Would the data collected by qualitative, quantitative discrete, or quantitative continuous?
Use the following information to answer the next two exercises: Thirty people spent two weeks around Mardi Gras in New Orleans. Their two-week weight gain is below. (Note: a loss is shown by a negative weight gain.)
| Weight Gain | Frequency |
|---|---|
| –2 | 3 |
| –1 | 5 |
| 0 | 2 |
| 1 | 4 |
| 4 | 13 |
| 6 | 2 |
| 11 | 1 |
7. Calculate the following values:
8. Construct a histogram and box plot of the data.
Use the following information to answer the next two exercises: A recent poll concerning credit cards found that 35 percent of respondents use a credit card that gives them a mile of air travel for every dollar they charge. Thirty percent of the respondents charge more than $2,000 per month. Of those respondents who charge more than $2,000, 80 percent use a credit card that gives them a mile of air travel for every dollar they charge.
9. What is the probability that a randomly selected respondent will spend more than $2,000 AND use a credit card that gives them a mile of air travel for every dollar they charge?
10. Are using a credit card that gives a mile of air travel for each dollar spent AND charging more than $2,000 per month independent events?
11. A sociologist wants to know the opinions of employed adult women about government funding for day care. She obtains a list of 520 members of a local business and professional women’s club and mails a questionnaire to 100 of these women selected at random. Sixty-eight questionnaires are returned. What is the population in this study?
Use the following information to answer the next two exercises: The next two questions refer to the following: An article from The San Jose Mercury News was concerned with the racial mix of the 1500 students at Prospect High School in Saratoga, CA. The table summarizes the results. (Male and female values are approximate.) Suppose one Prospect High School student is randomly selected.
| Gender/Ethnic group | White | Asian | Hispanic | Black | American Indian |
|---|---|---|---|---|---|
| Male | 400 | 468 | 115 | 35 | 16 |
| Female | 440 | 132 | 140 | 40 | 14 |
12. Find the probability that a student is Asian or Male.
13. Find the probability that a student is Black given that the student is female.
14. A sample of pounds lost, in a certain month, by individual members of a weight reducing clinic produced the following statistics:
15. What does it mean when a data set has a standard deviation equal to zero?
16. The statement that describe the illustration is:
17. According to a recent article in the San Jose Mercury News the average number of babies born with significant hearing loss (deafness) is approximately 2 per 1000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf.
18. A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.
Based upon the financial gain or loss over the long run, should you play the game?
Use the following information to answer the next four exercises: Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he/she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu.
19. Define the random variable and list its possible values.
20. State the distribution of X.
21. Find the probability that at least four of the 25 patients actually have the flu.
22. On average, for every 25 patients calling in, how many do you expect to have the flu?
Use the following information to answer the next two exercises: Different types of writing can sometimes be distinguished by the number of letters in the words used. A student interested in this fact wants to study the number of letters of words used by Tom Clancy in his novels. She opens a Clancy novel at random and records the number of letters of the first 250 words on the page.
23. What kind of data was collected?
24. What is the population under study?
Use the following information to answer the next seven exercises: A recent study of mothers of junior high school children in Santa Clara County reported that 76% of the mothers are employed in paid positions. Of those mothers who are employed, 64% work full-time (over 35 hours per week), and 36% work part-time. However, out of all of the mothers in the population, 49% work full-time. The population under study is made up of mothers of junior high school children in Santa Clara County. Let E = employed and F = full-time employment.
25.
26. The “type of employment” is considered to be what type of data?
27. Find the probability that a randomly selected mother works part-time given that she is employed.
28. Find the probability that a randomly selected person from the population will be employed or work full-time.
29. Being employed and working part-time:
Use the following additional information to answer the next two exercises: We randomly pick ten mothers from the above population. We are interested in the number of the mothers that are employed. Let X = number of mothers that are employed.
30. State the distribution for X.
31. Find the probability that at least six are employed.
32. We expect the statistics discussion board to have, on average, 14 questions posted to it per week. We are interested in the number of questions posted to it per day.
33. A person invests $1,000 into stock of a company that hopes to go public in one year. The probability that the person will lose all his money after one year (i.e. his stock will be worthless) is 35%. The probability that the person’s stock will still have a value of $1,000 after one year (i.e. no profit and no loss) is 60%. The probability that the person’s stock will increase in value by $10,000 after one year (i.e. will be worth $11,000) is 5%. Find the expected profit after one year.
34. Rachel’s piano cost $3,000. The average cost for a piano is $4,000 with a standard deviation of $2,500. Becca’s guitar cost $550. The average cost for a guitar is $500 with a standard deviation of $200. Matt’s drums cost $600. The average cost for drums is $700 with a standard deviation of $100. Whose cost was lowest when compared to his or her own instrument?
35. Explain why each statement is either true or false given the box plot in Table 13.45.
Using the following information to answer the next two exercises: 64 faculty members were asked the number of cars they owned (including spouse and children’s cars). The results are given in the following graph:
36. Find the approximate number of responses that were three.
37. Find the first, second and third quartiles. Use them to construct a box plot of the data.
Use the following information to answer the next three exercises: Table 13.45 shows data gathered from 15 girls on the Snow Leopard soccer team when they were asked how they liked to wear their hair. Supposed one girl from the team is randomly selected.
| Hair Style/Hair Color | Blond | Brown | Black |
|---|---|---|---|
| Ponytail | 3 | 2 | 5 |
| Plain | 2 | 2 | 1 |
38. Find the probability that the girl has black hair GIVEN that she wears a ponytail.
39. Find the probability that the girl wears her hair plain OR has brown hair.
40. Find the probability that the girl has blond hair AND that she wears her hair plain.
Use the following information to answer the next two exercises: X ~ U(3, 13)
41. Explain which of the following are false and which are true.
42. Calculate:
43. Which of the following is true for the box plot in Table 13.45?
44. If P(G|H) = P(G), then which of the following is correct?
45. If P(J) = 0.3, P(K) = 0.63, and J and K are independent events, then explain which are correct and which are incorrect.
46. On average, five students from each high school class get full scholarships to four-year colleges. Assume that most high school classes have about 500 students. X = the number of students from a high school class that get full scholarships to four-year schools. Which of the following is the distribution of X?
Use the following information to answer the next three exercises: Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery.
47. X ~ _________
48. The average wait time is:
49. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least 1.5 more hours is:
50. Given: X ~ Exp
51.
52. Which of the following distributions is described by the following example?
53. The length of time to brush one’s teeth is generally thought to be exponentially distributed with a mean of minutes. Find the probability that a randomly selected person brushes his or her teeth less than minutes.
54. Which distribution accurately describes the following situation?
55. A 2008 report on technology use states that approximately 20% of U.S. households have never sent an e-mail. Suppose that we select a random sample of fourteen U.S. households. Let X = the number of households in a 2008 sample of 14 households that have never sent an email
Use the following information to answer the next three exercises: Suppose that a sample of 15 randomly chosen people were put on a special weight loss diet. The amount of weight lost, in pounds, follows an unknown distribution with mean equal to 12 pounds and standard deviation equal to three pounds. Assume that the distribution for the weight loss is normal.
56. To find the probability that the mean amount of weight lost by 15 people is no more than 14 pounds, the random variable should be:
57. Find the probability asked for in Question 56.
58. Find the 90th percentile for the mean amount of weight lost by 15 people.
Using the following information to answer the next three exercises: The time of occurrence of the first accident during rush-hour traffic at a major intersection is uniformly distributed between the three hour interval 4 p.m. to 7 p.m. Let X = the amount of time (hours) it takes for the first accident to occur.
59. What is the probability that the time of occurrence is within the first half-hour or the last hour of the period from 4 to 7 p.m.?
60. The 20th percentile occurs after how many hours?
61. Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let C = the total cumulative time. Then C follows which distribution?
62. Using the information in Question 61, find the probability that the total time for all first accidents to occur is more than 43 hours.
Use the following information to answer the next two exercises: The length of time a parent must wait for his children to clean their rooms is uniformly distributed in the time interval from one to 15 days.
63. How long must a parent expect to wait for his children to clean their rooms?
64. What is the probability that a parent will wait more than six days given that the parent has already waited more than three days?
Use the following information to answer the next five exercises: Twenty percent of the students at a local community college live in within five miles of the campus. Thirty percent of the students at the same community college receive some kind of financial aid. Of those who live within five miles of the campus, 75% receive some kind of financial aid.
65. Find the probability that a randomly chosen student at the local community college does not live within five miles of the campus.
66. Find the probability that a randomly chosen student at the local community college lives within five miles of the campus or receives some kind of financial aid.
67. Are living in student housing within five miles of the campus and receiving some kind of financial aid mutually exclusive?
68. The interest rate charged on the financial aid is _______ data.
69. The following information is about the students who receive financial aid at the local community college.
Use the following information to answer the next two exercises: P(A) = 0.2, P(B) = 0.3; A and B are independent events.
70. P(A AND B) = ______
71. P(A OR B) = _______
72. If H and D are mutually exclusive events, P(H) = 0.25, P(D) = 0.15, then P(H|D).
73. Rebecca and Matt are 14 year old twins. Matt’s height is two standard deviations below the mean for 14 year old boys’ height. Rebecca’s height is 0.10 standard deviations above the mean for 14 year old girls’ height. Interpret this.
74. Construct a histogram of the IPO data (see Data Sets).
Use the following information to answer the next three exercises: Ninety homeowners were asked the number of estimates they obtained before having their homes fumigated. Let X = the number of estimates.
| x | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|
| 1 | 0.3 | |
| 2 | 0.2 | |
| 4 | 0.4 | |
| 5 | 0.1 |
75. Complete the cumulative frequency column.
76. Calculate the sample mean (a), the sample standard deviation (b) and the percent of the estimates that fall at or below four (c).
77. Calculate the median, M, the first quartile, Q1, the third quartile, Q3. Then construct a box plot of the data.
78. The middle 50% of the data are between _____ and _____.
Use the following information to answer the next three exercises: Seventy 5th and 6th graders were asked their favorite dinner.
| Pizza | Hamburgers | Spaghetti | Fried shrimp | |
|---|---|---|---|---|
| 5th grader | 15 | 6 | 9 | 0 |
| 6th grader | 15 | 7 | 10 | 8 |
79. Find the probability that one randomly chosen child is in the 6th grade and prefers fried shrimp.
80. Find the probability that a child does not prefer pizza.
81. Find the probability a child is in the 5th grade given that the child prefers spaghetti.
82. A sample of convenience is a random sample.
83. A statistic is a number that is a property of the population.
84. You should always throw out any data that are outliers.
85. Lee bakes pies for a small restaurant in Felton, CA. She generally bakes 20 pies in a day, on average. Of interest is the number of pies she bakes each day.
86. Six different brands of Italian salad dressing were randomly selected at a supermarket. The grams of fat per serving are 7, 7, 9, 6, 8, 5. Assume that the underlying distribution is normal. Calculate a 95% confidence interval for the population mean grams of fat per serving of Italian salad dressing sold in supermarkets.
87. Given: uniform, exponential, normal distributions. Match each to a statement below.
Use the following information to answer the next three exercises: In a survey at Kirkwood Ski Resort the following information was recorded:
| 0–10 | 11–20 | 21–40 | 40+ | |
|---|---|---|---|---|
| Ski | 10 | 12 | 30 | 8 |
| Snowboard | 6 | 17 | 12 | 5 |
Suppose that one person from Table 13.45 was randomly selected.
88. Find the probability that the person was a skier or was age 11–20.
89. Find the probability that the person was a snowboarder given he or she was age 21–40.
90. Explain which of the following are true and which are false.
91. The average length of time a person with a broken leg wears a cast is approximately six weeks. The standard deviation is about three weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the thirty people.
92. The distribution for X is uniform. What can we say for certain about the distribution for when n = 1?
93. The distribution for X is uniform. What can we say for certain about the distribution for when n = 50?
Use the following information to answer the next three exercises: A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots.
94. State the approximate distribution for , the distribution for the average lengths of baby carrots in 16 five-pound bags. ~ ______
95. Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches.
96. Find the probability that is between two and 2.25 inches.
Use the following information to answer the next three exercises: At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of five minutes and a standard deviation of two minutes.
97. Find the 90th percentile of waiting time in minutes.
98. Find the median waiting time for one student.
99. Find the probability that the average waiting time for 40 students is at least 4.5 minutes.
Use the following information to answer the next four exercises: Suppose that the time that owners keep their cars (purchased new) is normally distributed with a mean of seven years and a standard deviation of two years. We are interested in how long an individual keeps his car (purchased new). Our population is people who buy their cars new.
100. Sixty percent of individuals keep their cars at most how many years?
101. Suppose that we randomly survey one person. Find the probability that person keeps his or her car less than 2.5 years.
102. If we are to pick individuals ten at a time, find the distribution for the mean car length ownership.
103. If we are to pick ten individuals, find the probability that the sum of their ownership time is more than 55 years.
104. For which distribution is the median not equal to the mean?
105. Compare the standard normal distribution to the Student’s t-distribution, centered at zero. Explain which of the following are true and which are false.
Use the following information to answer the next five exercises: We are interested in the checking account balance of twenty-year-old college students. We randomly survey 16 twenty-year-old college students. We obtain a sample mean of $640 and a sample standard deviation of $150. Let X = checking account balance of an individual twenty year old college student.
106. Explain why we cannot determine the distribution of X.
107. If you were to create a confidence interval or perform a hypothesis test for the population mean checking account balance of twenty-year-old college students, what distribution would you use?
108. Find the 95% confidence interval for the true mean checking account balance of a twenty-year-old college student.
109. What type of data is the balance of the checking account considered to be?
110. What type of data is the number of twenty-year-olds considered to be?
111. On average, a busy emergency room gets a patient with a shotgun wound about once per week. We are interested in the number of patients with a shotgun wound the emergency room gets per 28 days.
Use the following information to answer the next two exercises: The probability that a certain slot machine will pay back money when a quarter is inserted is 0.30. Assume that each play of the slot machine is independent from each other. A person puts in 15 quarters for 15 plays.
112. Is the expected number of plays of the slot machine that will pay back money greater than, less than or the same as the median? Explain your answer.
113. Is it likely that exactly eight of the 15 plays would pay back money? Justify your answer numerically.
114. A game is played with the following rules:
115.
Use the following information to answer the next two exercises: A random sample of 70 compulsive gamblers were asked the number of days they go to casinos per week. The results are given in the following graph:
116. Find the number of responses that were five.
117. Find the mean, standard deviation, the median, the first quartile, the third quartile and the IQR.
118. Based upon research at De Anza College, it is believed that about 19% of the student population speaks a language other than English at home. Suppose that a study was done this year to see if that percent has decreased. Ninety-eight students were randomly surveyed with the following results. Fourteen said that they speak a language other than English at home.
119. Assume that you are an emergency paramedic called in to rescue victims of an accident. You need to help a patient who is bleeding profusely. The patient is also considered to be a high risk for contracting AIDS. Assume that the null hypothesis is that the patient does not have the HIV virus. What is a Type I error?
120. It is often said that Californians are more casual than the rest of Americans. Suppose that a survey was done to see if the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals. Fifty of each was surveyed with the following results. Fifteen Californians wear jeans to work and six non-Californians wear jeans to work.
Use the following information to answer the next two exercises: A group of Statistics students have developed a technique that they feel will lower their anxiety level on statistics exams. They measured their anxiety level at the start of the quarter and again at the end of the quarter. Recorded is the paired data in that order: (1000, 900); (1200, 1050); (600, 700); (1300, 1100); (1000, 900); (900, 900).
121. This is a test of (pick the best answer):
122. State the distribution to use for the test.
Use the following information to answer the next two exercises: A recent survey of U.S. teenage pregnancy was answered by 720 girls, age 12–19. Six percent of the girls surveyed said they have been pregnant. We are interested in the true proportion of U.S. girls, age 12–19, who have been pregnant.
123. Find the 95% confidence interval for the true proportion of U.S. girls, age 12–19, who have been pregnant.
124. The report also stated that the results of the survey are accurate to within ±3.7% at the 95% confidence level. Suppose that a new study is to be done. It is desired to be accurate to within 2% of the 95% confidence level. What is the minimum number that should be surveyed?
125. Given: X ~ Exp. Sketch the graph that depicts: P(x > 1).
Use the following information to answer the next three exercises: The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the mean amount of money a customer spends in one trip to the supermarket is $72.
126. Find the probability that one customer spends less than $72 in one trip to the supermarket?
127. Suppose five customers pool their money. How much money altogether would you expect the five customers to spend in one trip to the supermarket (in dollars)?
128. State the distribution to use if you want to find the probability that the mean amount spent by five customers in one trip to the supermarket is less than $60.
Use the following information to answer the next two exercises: Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the probability of water rationing is 10%. However, in any year, the probability of water rationing is 5%.
129. What is the probability of both a drought and water rationing occurring?
130. Out of the years with water rationing, find the probability that there is a drought.
Use the following information to answer the next three exercises:
| Apple | Pumpkin | Pecan | |
|---|---|---|---|
| Female | 40 | 10 | 30 |
| Male | 20 | 30 | 10 |
131. Suppose that one individual is randomly chosen. Find the probability that the person’s favorite pie is apple or the person is male.
132. Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan.
133. Conduct a hypothesis test to determine if favorite pie type and gender are independent.
Use the following information to answer the next two exercises: Let’s say that the probability that an adult watches the news at least once per week is 0.60.
134. We randomly survey 14 people. On average, how many people do we expect to watch the news at least once per week?
135. We randomly survey 14 people. Of interest is the number that watch the news at least once per week. State the distribution of X. X ~ _____
136. The following histogram is most likely to be a result of sampling from which distribution?
137. The ages of De Anza evening students is known to be normally distributed with a population mean of 40 and a population standard deviation of six. A sample of six De Anza evening students reported their ages (in years) as: 28; 35; 47; 45; 30; 50. Find the probability that the mean of six ages of randomly chosen students is less than 35 years. Hint: Find the sample mean.
138. A math exam was given to all the fifth grade children attending Country School. Two random samples of scores were taken. The null hypothesis is that the mean math scores for boys and girls in fifth grade are the same. Conduct a hypothesis test.
| n | s2 | ||
|---|---|---|---|
| Boys | 55 | 82 | 29 |
| Girls | 60 | 86 | 46 |
139. In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. Conduct a hypothesis test.
140. Which of the following is preferable when designing a hypothesis test?
Use the following information to answer the next three exercises: 120 people were surveyed as to their favorite beverage (non-alcoholic). The results are below.
| Beverage/Age | 0–9 | 10–19 | 20–29 | 30+ | Totals |
|---|---|---|---|---|---|
| Milk | 14 | 10 | 6 | 0 | 30 |
| Soda | 3 | 8 | 26 | 15 | 52 |
| Juice | 7 | 12 | 12 | 7 | 38 |
| Totals | 24 | 330 | 44 | 22 | 120 |
141. Are the events of milk and 30+:
142. Suppose that one person is randomly chosen. Find the probability that person is 10–19 given that he or she prefers juice.
143. Are “Preferred Beverage” and “Age” independent events? Conduct a hypothesis test.
144. Given the following histogram, which distribution is the data most likely to come from?
1. c. parameter
2. a. population
3. b. statistic
4. d. sample
5. e. variable
6. quantitative continuous
7.
8. Answers will vary.
9. c. (0.80)(0.30)
10. b. No, and they are not mutually exclusive either.
11. a. all employed adult women
12. 0.5773
13. 0.0522
14. b. The middle fifty percent of the members lost from 2 to 8.5 lbs.
15. c. All of the data have the same value.
16. c. The lowest data value is the median.
17. 0.279
18. b. No, I expect to come out behind in money.
19. X = the number of patients calling in claiming to have the flu, who actually have the flu.
20. B(25, 0.04)
21. 0.0165
22. 1
23. c. quantitative discrete
24. all words used by Tom Clancy in his novels
25.
26. qualitative
27. 0.36
28. 0.7636
29.
30. B(10, 0.76)
31. 0.9330
32.
33. $150
34. Matt
35.
36. 16
37. first quartile: 2
38. 0.5
39.
40.
41.
42.
43.
44. d. G and H are independent events.
45.
46. a. P(5)
47. a. U(0, 4)
48. b. 2 hour
49. a.
50.
51. c. 5 years
52. c. exponential
53. 0.63
54. B(14, 0.20)
55. B(14, 0.20)
56. c. the mean amount of weight lost by 15 people on the special weight loss diet.
57. 0.9951
58. 12.99
59. c.
60. b. 0.60
61. c. N(60, 5.477)
62. 0.9990
63. a. eight days
64. c. 0.7500
65. a. 80%
66. b. 35%
67. b. no
68. b. quantitative continuous
69. c. 150
70. d. 0.06
71. c. 0.44
72. b. 0
73. d. Matt is shorter than the average 14 year old boy.
74. Answers will vary.
75.
| x | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|
| 1 | 0.3 | 0.3 |
| 2 | 0.2 | 0.2 |
| 4 | 0.4 | 0.4 |
| 5 | 0.1 | 0.1 |
76.
77. M = 3; Q1 = 1; Q3 = 4
78. 1 and 4
79. d.
80. c.
81. a.
82. b. false
83. b. false
84. b. false
85.
86. CI: (5.25, 8.48)
87.
88.
89.
90.
91. N(180, 16.43)
92. a. The distribution for is still uniform with the same mean and standard deviation as the distribution for X.
93. c. The distribution for is normal with a larger mean and a larger standard deviation than the distribution for X.
94.
95. Answers will vary.
96. 0.5000
97. 7.6
98. 5
99. 0.9431
100. 7.5
101. 0.0122
102. N(7, 0.63)
103. 0.9911
104. b. Exponential
105.
106. Answers will vary.
107. Student’s t with df = 15
108. (560.07, 719.93)
109. quantitative continuous data
110. quantitative discrete data
111.
112. greater than
113. No; P(x = 8) = 0.0348
114. You will lose $5.
115. Becca
116. 14
117. Sample mean = 3.2
118. d. z = –1.19
119. We conclude that the patient does have the HIV virus when, in fact, the patient does not.
120. c. z = 2.21; p = 0.0136
121. c. dependent means
122. t5
123. (0.0424, 0.0770)
124. 2,401
125. Check student's solution.
126. 0.6321
127. $360
128.
129. 0.02
130. 0.40
131.
132.
133. p-value = 0; Reject the null hypothesis; conclude that they are dependent events
134. 8.4
135. B(14, 0.60)
136. d. Binomial
137. 0.3669
138. p-value = 0.0006; reject the null hypothesis; conclude that the averages are not equal
139. p-value = 0; reject the null hypothesis; conclude that the proportion of males is higher
140. Minimize α and β
141.
142.
143. No; p-value = 0
144. a. uniform
Data from the San Jose Mercury News.
Baran, Daya. “20 Percent of Americans Have Never Used Email.” Webguild.org, 2010. Available online at: http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed October 17, 2013).
Data from Parade Magazine.
Use the following information to answer the next three exercises. A grocery store is interested in how much money, on average, their customers spend each visit in the produce department. Using their store records, they draw a sample of 1,000 visits and calculate each customer’s average spending on produce.
1. Identify the population, sample, parameter, statistic, variable, and data for this example.
2. What kind of data is “amount of money spent on produce per visit”?
3. The study finds that the mean amount spent on produce per visit by the customers in the sample is $12.84. This is an example of a:
Use the following information to answer the next two exercises. A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a short survey including information about how many times they have visited the club in the past week.
4. What kind of a sampling design is this?
5. “Number of visits per week” is what kind of data?
6. Describe a situation in which you would calculate a parameter, rather than a statistic.
7. The U.S. federal government conducts a survey of high school seniors concerning their plans for future education and employment. One question asks whether they are planning to attend a four-year college or university in the following year. Fifty percent answer yes to this question; that fifty percent is a:
8. Imagine that the U.S. federal government had the means to survey all high school seniors in the U.S. concerning their plans for future education and employment, and found that 50 percent were planning to attend a 4-year college or university in the following year. This 50 percent is an example of a:
Use the following information to answer the next three exercises. A survey of a random sample of 100 nurses working at a large hospital asked how many years they had been working in the profession. Their answers are summarized in the following (incomplete) table.
9. Fill in the blanks in the table and round your answers to two decimal places for the Relative Frequency and Cumulative Relative Frequency cells.
| # of years | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| < 5 | 25 | ||
| 5–10 | 30 | ||
| > 10 | empty |
10. What proportion of nurses have five or more years of experience?
11. What proportion of nurses have ten or fewer years of experience?
12. Describe how you might draw a random sample of 30 students from a lecture class of 200 students.
13. Describe how you might draw a stratified sample of students from a college, where the strata are the students’ class standing (freshman, sophomore, junior, or senior).
14. A manager wants to draw a sample, without replacement, of 30 employees from a workforce of 150. Describe how the chance of being selected will change over the course of drawing the sample.
15. The manager of a department store decides to measure employee satisfaction by selecting four departments at random, and conducting interviews with all the employees in those four departments. What type of survey design is this?
16. A popular American television sports program conducts a poll of viewers to see which team they believe will win the NFL (National Football League) championship this year. Viewers vote by calling a number displayed on the television screen and telling the operator which team they think will win. Do you think that those who participate in this poll are representative of all football fans in America?
17. Two researchers studying vaccination rates independently draw samples of 50 children, ages 3–18 months, from a large urban area, and determine if they are up to date on their vaccinations. One researcher finds that 84 percent of the children in her sample are up to date, and the other finds that 86 percent in his sample are up to date. Assuming both followed proper sampling procedures and did their calculations correctly, what is a likely explanation for this discrepancy?
18. A high school increased the length of the school day from 6.5 to 7.5 hours. Students who wished to attend this high school were required to sign contracts pledging to put forth their best effort on their school work and to obey the school rules; if they did not wish to do so, they could attend another high school in the district. At the end of one year, student performance on statewide tests had increased by ten percentage points over the previous year. Does this improvement prove that a longer school day improves student achievement?
19. You read a newspaper article reporting that eating almonds leads to increased life satisfaction. The study was conducted by the Almond Growers Association, and was based on a randomized survey asking people about their consumption of various foods, including almonds, and also about their satisfaction with different aspects of their life. Does anything about this poll lead you to question its conclusion?
20. Why is non-response a problem in surveys?
21. Compute the mean of the following numbers, and report your answer using one more decimal place than is present in the original data:
22. A psychologist is interested in whether the size of tableware (bowls, plates, etc.) influences how much college students eat. He randomly assigns 100 college students to one of two groups: the first is served a meal using normal-sized tableware, while the second is served the same meal, but using tableware that it 20 percent smaller than normal. He records how much food is consumed by each group. Identify the following components of this study.
23. A researcher analyzes the results of the SAT (Scholastic Aptitude Test) over a five-year period and finds that male students on average score higher on the math section, and female students on average score higher on the verbal section. She concludes that these observed differences in test performance are due to genetic factors. Explain how lurking variables could offer an alternative explanation for the observed differences in test scores.
24. Explain why it would not be possible to use random assignment to study the health effects of smoking.
25. A professor conducts a telephone survey of a city’s population by drawing a sample of numbers from the phone book and having her student assistants call each of the selected numbers once to administer the survey. What are some sources of bias with this survey?
26. A professor offers extra credit to students who take part in her research studies. What is an ethical problem with this method of recruiting subjects?
Use the following information to answer the next four exercises. The midterm grades on a chemistry exam, graded on a scale of 0 to 100, were:
27. Do you see any outliers in this data? If so, how would you address the situation?
28. Construct a stem plot for this data, using only the values in the range 0–100.
29. Describe the distribution of exam scores.
30. In a class of 35 students, seven students received scores in the 70–79 range. What is the relative frequency of scores in this range?
Use the following information to answer the next three exercises. You conduct a poll of 30 students to see how many classes they are taking this term. Your results are:
31. You decide to construct a histogram of this data. What will be the range of your first bar, and what will be the central point?
32. What will be the widths and central points of the other bars?
33. Which bar in this histogram will be the tallest, and what will be its height?
34. You get data from the U.S. Census Bureau on the median household income for your city, and decide to display it graphically. Which is the better choice for this data, a bar graph or a histogram?
35. You collect data on the color of cars driven by students in your statistics class, and want to display this information graphically. Which is the better choice for this data, a bar graph or a histogram?
36. Your daughter brings home test scores showing that she scored in the 80th percentile in math and the 76th percentile in reading for her grade. Interpret these scores.
37. You have to wait 90 minutes in the emergency room of a hospital before you can see a doctor. You learn that your wait time was in the 82nd percentile of all wait times. Explain what this means, and whether you think it is good or bad.
Use the following information to answer the next three exercises. 1; 1; 2; 3; 4; 4; 5; 5; 6; 7; 7; 8; 9
38. What is the median for this data?
39. What is the first quartile for this data?
40. What is the third quartile for this data?
Use the following information to answer the next four exercises. This box plot represents scores on the final exam for a physics class.
41. What is the median for this data, and how do you know?
42. What are the first and third quartiles for this data, and how do you know?
43. What is the interquartile range for this data?
44. What is the range for this data?
45. In a marathon, the median finishing time was 3:35:04 (three hours, 35 minutes, and four seconds). You finished in 3:34:10. Interpret the meaning of the median time, and discuss your time in relation to it.
Use the following information to answer the next three exercises. The value, in thousands of dollars, for houses on a block, are: 45; 47; 47.5; 51; 53.5; 125.
46. Calculate the mean for this data.
47. Calculate the median for this data.
48. Which do you think better reflects the average value of the homes on this block?
49. In a left-skewed distribution, which is greater?
50. In a right-skewed distribution, which is greater?
51. In a symmetrical distribution what will be the relationship among the mean, median, and mode?
Use the following information to answer the next four exercises. 10; 11; 15; 15; 17; 22
52. Compute the mean and standard deviation for this data; use the sample formula for the standard deviation.
53. What number is two standard deviations above the mean of this data?
54. Express the number 13.7 in terms of the mean and standard deviation of this data.
55. In a biology class, the scores on the final exam were normally distributed, with a mean of 85, and a standard deviation of five. Susan got a final exam score of 95. Express her exam result as a z-score, and interpret its meaning.
Use the following information to answer the next two exercises. You have a jar full of marbles: 50 are red, 25 are blue, and 15 are yellow. Assume you draw one marble at random for each trial, and replace it before the next trial.
56. Find P(B).
57. Which is more likely, drawing a red marble or a yellow marble? Justify your answer numerically.
Use the following information to answer the next two exercises. The following are probabilities describing a group of college students.
58. Write the symbols for the probability that a student, selected at random, is both female and a science major.
59. Write the symbols for the probability that the student is an education major, given that the student is male.
60. Events A and B are independent.
61. C and D are mutually exclusive events.
62. In a high school graduating class of 300, 200 students are going to college, 40 are planning to work full-time, and 80 are taking a gap year. Are these events mutually exclusive?
Use the following information to answer the next two exercises. An archer hits the center of the target (the bullseye) 70 percent of the time. However, she is a streak shooter, and if she hits the center on one shot, her probability of hitting it on the shot immediately following is 0.85. Written in probability notation:
63. Calculate the probability that she will hit the center of the target on two consecutive shots.
64. Are P(A) and P(B) independent in this example?
Use the following information to answer the next three exercises. The following contingency table displays the number of students who report studying at least 15 hours per week, and how many made the honor roll in the past semester.
| Honor roll | No honor roll | Total | |
|---|---|---|---|
| Study at least 15 hours/week | 200 | ||
| Study less than 15 hours/week | 125 | 193 | |
| Total | 1,000 |
65. Complete the table.
66. Find P(honor roll|study at least 15 hours per week).
67. What is the probability a student studies less than 15 hours per week?
68. Are the events “study at least 15 hours per week” and “makes the honor roll” independent? Justify your answer numerically.
69. At a high school, some students play on the tennis team, some play on the soccer team, but neither plays both tennis and soccer. Draw a Venn diagram illustrating this.
70. At a high school, some students play tennis, some play soccer, and some play both. Draw a Venn diagram illustrating this.
1.
2. c
3. d
4. d
5. c
6. Answers will vary.
7. b
8. a
9.
| # of years | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| < 5 | 25 | 0.25 | 0.25 |
| 5–10 | 30 | 0.30 | 0.55 |
| > 10 | 45 | 0.45 | 1.00 |
10. 0.75
11. 0.55
12. Answers will vary.
13. One possibility would be to obtain a roster of students enrolled in the college, including the class standing for each student. Then you would draw a proportionate random sample from within each class (for instance, if 30 percent of the students in the college are freshman, then 30 percent of your sample would be drawn from the freshman class).
14. For the first person picked, the chance of any individual being selected is one in 150. For the second person, it is one in 149, for the third it is one in 148, and so on. For the 30th person selected, the chance of selection is one in 121.
15. a
16. No. There are at least two chances for bias. First, the viewers of this particular program may not be representative of American football fans as a whole. Second, the sample will be self-selected, because people have to make a phone call in order to take part, and those people are probably not representative of the American football fan population as a whole.
17. These results (84 percent in one sample, 86 percent in the other) are probably due to sampling variability. Each researcher drew a different sample of children, and you would not expect them to get exactly the same result, although you would expect the results to be similar, as they are in this case.
18. No. The improvement could also be due to self-selection: only motivated students were willing to sign the contract, and they would have done well even in a school with 6.5 hour days. Because both changes were implemented at the same time, it is not possible to separate out their influence.
19. At least two aspects of this poll are troublesome. The first is that it was conducted by a group who would benefit by the result—almond sales are likely to increase if people believe that eating almonds will make them happier. The second is that this poll found that almond consumption and life satisfaction are correlated, but does not establish that eating almonds causes satisfaction. It is equally possible, for instance, that people with higher incomes are more likely to eat almonds, and are also more satisfied with their lives.
20. You want the sample of people who take part in a survey to be representative of the population from which they are drawn. People who refuse to take part in a survey often have different views than those who do participate, and so even a random sample may produce biased results if a large percentage of those selected refuse to participate in a survey.
21. 13.2
22.
23. There are many lurking variables that could influence the observed differences in test scores. Perhaps the boys, on average, have taken more math courses than the girls, and the girls have taken more English classes than the boys. Perhaps the boys have been encouraged by their families and teachers to prepare for a career in math and science, and thus have put more effort into studying math, while the girls have been encouraged to prepare for fields like communication and psychology that are more focused on language use. A study design would have to control for these and other potential lurking variables (anything that could explain the observed difference in test scores, other than the genetic explanation) in order to draw a scientifically sound conclusion about genetic differences.
24. To use random assignment, you would have to be able to assign people to either smoke or not smoke. Because smoking has many harmful effects, this would not be an ethical experiment. Instead, we study people who have chosen to smoke, and compare them to others who have chosen not to smoke, and try to control for the other ways those two groups may differ (lurking variables).
25. Sources of bias include the fact that not everyone has a telephone, that cell phone numbers are often not listed in published directories, and that an individual might not be at home at the time of the phone call; all these factors make it likely that the respondents to the survey will not be representative of the population as a whole.
26. Research subjects should not be coerced into participation, and offering extra credit in exchange for participation could be construed as coercion. In addition, this method will result in a volunteer sample, which cannot be assumed to be representative of the population as a whole.
27. The value 740 is an outlier, because the exams were graded on a scale of 0 to 100, and 740 is far outside that range. It may be a data entry error, with the actual score being 74, so the professor should check that exam again to see what the actual score was.
28.
| Stem | Leaf |
|---|---|
| 6 | 2 4 5 5 8 |
| 7 | 0 2 2 4 5 5 5 6 8 8 |
| 8 | 1 3 3 4 5 7 8 |
| 9 | 2 5 8 8 |
| 10 | 0 0 |
29. Most scores on this exam were in the range of 70–89, with a few scoring in the 60–69 range, and a few in the 90–100 range.
30.
31. The range will be 0.5–1.5, and the central point will be 1.
32. Range 1.5–2.5, central point 2; range 2.5–3.5, central point 3; range 3.5–4.5, central point 4; range 4.5–5.5., central point 5.
33. The bar from 3.5 to 4.5, with a central point of 4, will be tallest; its height will be nine, because there are nine students taking four courses.
34. The histogram is a better choice, because income is a continuous variable.
35. A bar graph is the better choice, because this data is categorical rather than continuous.
36. Your daughter scored better than 80 percent of the students in her grade on math and better than 76 percent of the students in reading. Both scores are very good, and place her in the upper quartile, but her math score is slightly better in relation to her peers than her reading score.
37. You had an unusually long wait time, which is bad: 82 percent of patients had a shorter wait time than you, and only 18 percent had a longer wait time.
38. 5
39. 3
40. 7
41. The median is 86, as represented by the vertical line in the box.
42. The first quartile is 80, and the third quartile is 92, as represented by the left and right boundaries of the box.
43. IQR = 92 – 80 = 12
44. Range = 100 – 75 = 25
45. Half the runners who finished the marathon ran a time faster than 3:35:04, and half ran a time slower than 3:35:04. Your time is faster than the median time, so you did better than more than half of the runners in this race.
46. 61.5, or $61,500
47. 49.25 or $49,250
48. The median, because the mean is distorted by the high value of one house.
49. c
50. a
51. They will all be fairly close to each other.
52. Mean: 15
53. 15 + (2)(4.3) = 23.6
54. 13.7 is one standard deviation below the mean of this data, because 15 – 4.3 = 10.7
55.
56.
57. Drawing a red marble is more likely.
58. P(F AND S)
59. P(E|M)
60. P(A AND B) = (0.3)(0.5) = 0.15
61. P(C OR D) = 0.18 + 0.03 = 0.21
62. No, they cannot be mutually exclusive, because they add up to more than 300. Therefore, some students must fit into two or more categories (e.g., both going to college and working full time).
63. P(A and B) = (P(B|A))(P(A)) = (0.85)(0.70) = 0.595
64. No. If they were independent, P(B) would be the same as P(B|A). We know this is not the case, because P(B) = 0.70 and P(B|A) = 0.85.
65.
| Honor roll | No honor roll | Total | |
|---|---|---|---|
| Study at least 15 hours/week | 482 | 200 | 682 |
| Study less than 15 hours/week | 125 | 193 | 318 |
| Total | 607 | 393 | 1,000 |
66.
67.
68. Let P(S) = study at least 15 hours per week
69.
70.
Use the following information to answer the next five exercises. You conduct a survey among a random sample of students at a particular university. The data collected includes their major, the number of classes they took the previous semester, and amount of money they spent on books purchased for classes in the previous semester.
1. If X = student’s major, then what is the domain of X?
2. If Y = the number of classes taken in the previous semester, what is the domain of Y?
3. If Z = the amount of money spent on books in the previous semester, what is the domain of Z?
4. Why are X, Y, and Z in the previous example random variables?
5. After collecting data, you find that for one case, z = –7. Is this a possible value for Z?
6. What are the two essential characteristics of a discrete probability distribution?
Use this discrete probability distribution represented in this table to answer the following six questions. The university library records the number of books checked out by each patron over the course of one day, with the following result:
| x | P(x) |
|---|---|
| 0 | 0.20 |
| 1 | 0.45 |
| 2 | 0.20 |
| 3 | 0.10 |
| 4 | 0.05 |
7. Define the random variable X for this example.
8. What is P(x > 2)?
9. What is the probability that a patron will check out at least one book?
10. What is the probability a patron will take out no more than three books?
11. If the table listed P(x) as 0.15, how would you know that there was a mistake?
12. What is the average number of books taken out by a patron?
Use the following information to answer the next four exercises. Three jobs are open in a company: one in the accounting department, one in the human resources department, and one in the sales department. The accounting job receives 30 applicants, and the human resources and sales department 60 applicants.
13. If X = the number of applications for a job, use this information to fill in Table 13.45.
| x | P(x) | xP(x) |
|---|---|---|
14. What is the mean number of applicants?
15. What is the PDF for X?
16. Add a fourth column to the table, for (x – μ)2P(x).
17. What is the standard deviation of X?
18. In a binomial experiment, if p = 0.65, what does q equal?
19. What are the required characteristics of a binomial experiment?
20. Joe conducts an experiment to see how many times he has to flip a coin before he gets four heads in a row. Does this qualify as a binomial experiment?
Use the following information to answer the next three exercises. In a particularly community, 65 percent of households include at least one person who has graduated from college. You randomly sample 100 households in this community. Let X = the number of households including at least one college graduate.
21. Describe the probability distribution of X.
22. What is the mean of X?
23. What is the standard deviation of X?
Use the following information to answer the next four exercises. Joe is the star of his school’s baseball team. His batting average is 0.400, meaning that for every ten times he comes to bat (an at-bat), four of those times he gets a hit. You decide to track his batting performance his next 20 at-bats.
24. Define the random variable X in this experiment.
25. Assuming Joe’s probability of getting a hit is independent and identical across all 20 at-bats, describe the distribution of X.
26. Given this information, what number of hits do you predict Joe will get?
27. What is the standard deviation of X?
28. What are the three major characteristics of a geometric experiment?
29. You decide to conduct a geometric experiment by flipping a coin until it comes up heads. This takes five trials. Represent the outcomes of this trial, using H for heads and T for tails.
30. You are conducting a geometric experiment by drawing cards from a normal 52-card pack, with replacement, until you draw the Queen of Hearts. What is the domain of X for this experiment?
31. You are conducting a geometric experiment by drawing cards from a normal 52-card deck, without replacement, until you draw a red card. What is the domain of X for this experiment?
Use the following information to answer the next three exercises. In a particular university, 27 percent of students are engineering majors. You decide to select students at random until you choose one that is an engineering major. Let X = the number of students you select until you find one that is an engineering major.
32. What is the probability distribution of X?
33. What is the mean of X?
34. What is the standard deviation of X?
35. You draw a random sample of ten students to participate in a survey, from a group of 30, consisting of 16 boys and 14 girls. You are interested in the probability that seven of the students chosen will be boys. Does this qualify as a hypergeometric experiment? List the conditions and whether or not they are met.
36. You draw five cards, without replacement, from a normal 52-card deck of playing cards, and are interested in the probability that two of the cards are spades. What are the group of interest, size of the group of interest, and sample size for this example?
37. What are the key characteristics of the Poisson distribution?
Use the following information to answer the next three exercises. The number of drivers to arrive at a toll booth in an hour can be modeled by the Poisson distribution.
38. If X = the number of drivers, and the average numbers of drivers per hour is four, how would you express this distribution?
39. What is the domain of X?
40. What are the mean and standard deviation of X?
41. You conduct a survey of students to see how many books they purchased the previous semester, the total amount they paid for those books, the number they sold after the semester was over, and the amount of money they received for the books they sold. Which variables in this survey are discrete, and which are continuous?
42. With continuous random variables, we never calculate the probability that X has a particular value, but always speak in terms of the probability that X has a value within a particular range. Why is this?
43. For a continuous random variable, why are P(x < c) and P(x ≤ c) equivalent statements?
44. For a continuous probability function, P(x < 5) = 0.35. What is P(x > 5), and how do you know?
45. Describe how you would draw the continuous probability distribution described by the function for . What type of a distribution is this?
46. For the continuous probability distribution described by the function for , what is the P(0 < x < 4)?
47. For the continuous probability distribution described by the function for , what is the P(2 < x < 5)?
Use the following information to answer the next four exercises. The number of minutes that a patient waits at a medical clinic to see a doctor is represented by a uniform distribution between zero and 30 minutes, inclusive.
48. If X equals the number of minutes a person waits, what is the distribution of X?
49. Write the probability density function for this distribution.
50. What is the mean and standard deviation for waiting time?
51. What is the probability that a patient waits less than ten minutes?
52. The distribution of the variable X, representing the average time to failure for an automobile battery, can be written as: X ~ Exp(m). Describe this distribution in words.
53. If the value of m for an exponential distribution is ten, what are the mean and standard deviation for the distribution?
54. Write the probability density function for a variable distributed as: X ~ Exp(0.2).
55. Translate this statement about the distribution of a random variable X into words: X ~ (100, 15).
56. If the variable X has the standard normal distribution, express this symbolically.
Use the following information for the next six exercises. According to the World Health Organization, distribution of height in centimeters for girls aged five years and no months has the distribution: X ~ N(109, 4.5).
57. What is the z-score for a height of 112 inches?
58. What is the z-score for a height of 100 centimeters?
59. Find the z-score for a height of 105 centimeters and explain what that means In the context of the population.
60. What height corresponds to a z-score of 1.5 in this population?
61. Using the empirical rule, we expect about 68 percent of the values in a normal distribution to lie within one standard deviation above or below the mean. What does this mean, in terms of a specific range of values, for this distribution?
62. Using the empirical rule, about what percent of heights in this distribution do you expect to be between 95.5 cm and 122.5 cm?
Use the following information to answer the next four exercises. The distributor of lotto tickets claims that 20 percent of the tickets are winners. You draw a sample of 500 tickets to test this proposition.
63. Can you use the normal approximation to the binomial for your calculations? Why or why not.
64. What are the expected mean and standard deviation for your sample, assuming the distributor’s claim is true?
65. What is the probability that your sample will have a mean greater than 100?
66. If the z-score for your sample result is –2.00, explain what this means, using the empirical rule.
67. What does the central limit theorem state with regard to the distribution of sample means?
68. The distribution of results from flipping a fair coin is uniform: heads and tails are equally likely on any flip, and over a large number of trials, you expect about the same number of heads and tails. Yet if you conduct a study by flipping 30 coins and recording the number of heads, and repeat this 100 times, the distribution of the mean number of heads will be approximately normal. How is this possible?
69. The mean of a normally-distributed population is 50, and the standard deviation is four. If you draw 100 samples of size 40 from this population, describe what you would expect to see in terms of the sampling distribution of the sample mean.
70. X is a random variable with a mean of 25 and a standard deviation of two. Write the distribution for the sample mean of samples of size 100 drawn from this population.
71. Your friend is doing an experiment drawing samples of size 50 from a population with a mean of 117 and a standard deviation of 16. This sample size is large enough to allow use of the central limit theorem, so he says the standard deviation of the sampling distribution of sample means will also be 16. Explain why this is wrong, and calculate the correct value.
72. You are reading a research article that refers to “the standard error of the mean.” What does this mean, and how is it calculated?
Use the following information to answer the next six exercises. You repeatedly draw samples of n = 100 from a population with a mean of 75 and a standard deviation of 4.5.
73. What is the expected distribution of the sample means?
74. One of your friends tries to convince you that the standard error of the mean should be 4.5. Explain what error your friend made.
75. What is the z-score for a sample mean of 76?
76. What is the z-score for a sample mean of 74.7?
77. What sample mean corresponds to a z-score of 1.5?
78. If you decrease the sample size to 50, will the standard error of the mean be smaller or larger? What would be its value?
Use the following information to answer the next two questions. We use the empirical rule to analyze data for samples of size 60 drawn from a population with a mean of 70 and a standard deviation of 9.
79. What range of values would you expect to include 68 percent of the sample means?
80. If you increased the sample size to 100, what range would you expect to contain 68 percent of the sample means, applying the empirical rule?
81. How does the central limit theorem apply to sums of random variables?
82. Explain how the rules applying the central limit theorem to sample means, and to sums of a random variable, are similar.
83. If you repeatedly draw samples of size 50 from a population with a mean of 80 and a standard deviation of four, and calculate the sum of each sample, what is the expected distribution of these sums?
Use the following information to answer the next four exercises. You draw one sample of size 40 from a population with a mean of 125 and a standard deviation of seven.
84. Compute the sum. What is the probability that the sum for your sample will be less than 5,000?
85. If you drew samples of this size repeatedly, computing the sum each time, what range of values would you expect to contain 95 percent of the sample sums?
86. What value is one standard deviation below the mean?
87. What value corresponds to a z-score of 2.2?
88. What does the law of large numbers say about the relationship between the sample mean and the population mean?
89. Applying the law of large numbers, which sample mean would expect to be closer to the population mean, a sample of size ten or a sample of size 100?
Use this information for the next three questions. A manufacturer makes screws with a mean diameter of 0.15 cm (centimeters) and a range of 0.10 cm to 0.20 cm; within that range, the distribution is uniform.
90. If X = the diameter of one screw, what is the distribution of X?
91. Suppose you repeatedly draw samples of size 100 and calculate their mean. Applying the central limit theorem, what is the distribution of these sample means?
92. Suppose you repeatedly draw samples of 60 and calculate their sum. Applying the central limit theorem, what is the distribution of these sample sums?
1. The domain of X = {English, Mathematics,….], i.e., a list of all the majors offered at the university, plus “undeclared.”
2. The domain of Y = {0, 1, 2, …}, i.e., the integers from 0 to the upper limit of classes allowed by the university.
3. The domain of Z = any amount of money from 0 upwards.
4. Because they can take any value within their domain, and their value for any particular case is not known until the survey is completed.
5. No, because the domain of Z includes only positive numbers (you can’t spend a negative amount of money). Possibly the value –7 is a data entry error, or a special code to indicated that the student did not answer the question.
6. The probabilities must sum to 1.0, and the probabilities of each event must be between 0 and 1, inclusive.
7. Let X = the number of books checked out by a patron.
8. P(x > 2) = 0.10 + 0.05 = 0.15
9. P(x ≥ 0) = 1 – 0.20 = 0.80
10. P(x ≤ 3) = 1 – 0.05 = 0.95
11. The probabilities would sum to 1.10, and the total probability in a distribution must always equal 1.0.
12. = 0(0.20) + 1(0.45) + 2(0.20) + 3(0.10) + 4(0.05) = 1.35
13.
| x | P(x) | xP(x) |
|---|---|---|
| 30 | 0.33 | 9.90 |
| 40 | 0.33 | 13.20 |
| 60 | 0.33 | 19.80 |
14. = 9.90 + 13.20 + 19.80 = 42.90
15. P(x = 30) = 0.33
16.
| x | P(x) | xP(x) | (x – μ)2P(x) |
|---|---|---|---|
| 30 | 0.33 | 9.90 | (30 – 42.90)2(0.33) = 54.91 |
| 40 | 0.33 | 13.20 | (40 – 42.90)2(0.33) = 2.78 |
| 60 | 0.33 | 19.90 | (60 – 42.90)2(0.33) = 96.49 |
17.
18. q = 1 – 0.65 = 0.35
19.
20. No, because there are not a fixed number of trials
21. X ~ B(100, 0.65)
22. μ = np = 100(0.65) = 65
23.
24. X = Joe gets a hit in one at-bat (in one occasion of his coming to bat)
25. X ~ B(20, 0.4)
26. μ = np = 20(0.4) = 8
27.
28.
29. T T T T H
30. The domain of X = {1, 2, 3, 4, 5, ….n}. Because you are drawing with replacement, there is no upper bound to the number of draws that may be necessary.
31. The domain of X = {1, 2, 3, 4, 5, 6, 7, 8., 9, 10, 11, 12…27}. Because you are drawing without replacement, and 26 of the 52 cards are red, you have to draw a red card within the first 17 draws.
32. X ~ G(0.24)
33.
34.
35. Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials).
36. The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five.
37. A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known.
38. X ~ P(4)
39. The domain of X = {0, 1, 2, 3, …..) i.e., any integer from 0 upwards.
40.
41. The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold.
42. Because for a continuous random variable, P(x = c) = 0, where c is any single value. Instead, we calculate P(c < x < d), i.e., the probability that the value of x is between the values c and d.
43. Because P(x = c) = 0 for any continuous random variable.
44. P(x > 5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1.
45. This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at and 0.
46.
47.
48. X ~ U(0, 15)
49. for for
50.
51.
52. X has an exponential distribution with decay parameter m and mean and standard deviation . In this distribution, there will be a relatively large numbers of small values, with values becoming less common as they become larger.
53.
54. f(x) = 0.2e–0.2x where x ≥ 0.
55. The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15.
56. X ~ N(0,1)
57. so
58. so
59.
60. 109 + (1.5)(4.5) = 115.75 cm
61. We expect about 68 percent of the heights of girls of age five years and zero months to be between 104.5 cm and 113.5 cm.
62. We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean.
63. Yes, because both np and nq are greater than five.
64.
65. Fifty percent, because in a normal distribution, half the values lie above the mean.
66. The results of our sample were two standard deviations below the mean, suggesting it is unlikely that 20 percent of the lotto tickets are winners, as claimed by the distributor, and that the true percent of winners is lower. Applying the Empirical Rule, If that claim were true, we would expect to see a result this far below the mean only about 2.5 percent of the time.
67. The central limit theorem states that if samples of sufficient size drawn from a population, the distribution of sample means will be normal, even if the distribution of the population is not normal.
68. The sample size of 30 is sufficiently large in this example to apply the central limit theorem. This theorem ] states that for samples of sufficient size drawn from a population, the sampling distribution of the sample mean will approach normality, regardless of the distribution of the population from which the samples were drawn.
69. You would not expect each sample to have a mean of 50, because of sampling variability. However, you would expect the sampling distribution of the sample means to cluster around 50, with an approximately normal distribution, so that values close to 50 are more common than values further removed from 50.
70. because
71. The standard deviation of the sampling distribution of the sample means can be calculated using the formula , which in this case is . The correct value for the standard deviation of the sampling distribution of the sample means is therefore 2.26.
72. The standard error of the mean is another name for the standard deviation of the sampling distribution of the sample mean. Given samples of size n drawn from a population with standard deviation σx, the standard error of the mean is .
73. X ~ N(75, 0.45)
74. Your friend forgot to divide the standard deviation by the square root of n.
75.
76.
77. 75 + (1.5)(0.45) = 75.675
78. The standard error of the mean will be larger, because you will be dividing by a smaller number. The standard error of the mean for samples of size n = 50 is:
79. You would expect this range to include values up to one standard deviation above or below the mean of the sample means. In this case:
80. and so you would expect 68 percent of the sample means to be between 69.1 and 70.9. Note that this is a narrower interval due to the increased sample size.
81. For a random variable X, the random variable ΣX will tend to become normally distributed as the size n of the samples used to compute the sum increases.
82. Both rules state that the distribution of a quantity (the mean or the sum) calculated on samples drawn from a population will tend to have a normal distribution, as the sample size increases, regardless of the distribution of population from which the samples are drawn.
83. so
84.The probability is 0.50, because 5,000 is the mean of the sampling distribution of sums of size 40 from this population. Sums of random variables computed from a sample of sufficient size are normally distributed, and in a normal distribution, half the values lie below the mean.
85. Using the empirical rule, you would expect 95 percent of the values to be within two standard deviations of the mean. Using the formula for the standard deviation is for a sample sum: so you would expect 95 percent of the values to be between 5,000 + (2)(44.3) and 5,000 – (2)(44.3), or between 4,911.4 and 588.6.
86.
87.
88. The law of large numbers says that as sample size increases, the sample mean tends to get nearer and nearer to the population mean.
89. You would expect the mean from a sample of size 100 to be nearer to the population mean, because the law of large numbers says that as sample size increases, the sample mean tends to approach the population mea.
90. X ~ N(0.10, 0.20)
91. and the standard deviation of a uniform distribution is . In this example, the standard deviation of the distribution is
92.
Use the following information to answer the next seven exercises. You draw a sample of size 30 from a normally distributed population with a standard deviation of four.
1. What is the standard error of the sample mean in this scenario, rounded to two decimal places?
2. What is the distribution of the sample mean?
3. If you want to construct a two-sided 95% confidence interval, how much probability will be in each tail of the distribution?
4. What is the appropriate z-score and error bound or margin of error (EBM) for a 95% confidence interval for this data?
5. Rounding to two decimal places, what is the 95% confidence interval if the sample mean is 41?
6. What is the 90% confidence interval if the sample mean is 41? Round to two decimal places
7. Suppose the sample size in this study had been 50, rather than 30. What would the 95% confidence interval be if the sample mean is 41? Round your answer to two decimal places.
8. For any given data set and sampling situation, which would you expect to be wider: a 95% confidence interval or a 99% confidence interval?
9. Comparing graphs of the standard normal distribution (z-distribution) and a t-distribution with 15 degrees of freedom (df), how do they differ?
10. Comparing graphs of the standard normal distribution (z-distribution) and a t-distribution with 15 degrees of freedom (df), how are they similar?
Use the following information to answer the next five exercises. Body temperature is known to be distributed normally among healthy adults. Because you do not know the population standard deviation, you use the t-distribution to study body temperature. You collect data from a random sample of 20 healthy adults and find that your sample temperatures have a mean of 98.4 and a sample standard deviation of 0.3 (both in degrees Fahrenheit).
11. What is the degrees of freedom (df) for this study?
12. For a two-tailed 95% confidence interval, what is the appropriate t-value to use in the formula?
13. What is the 95% confidence interval?
14. What is the 99% confidence interval? Round to two decimal places.
15. Suppose your sample size had been 30 rather than 20. What would the 95% confidence interval be then? Round to two decimal places
Use this information to answer the next four exercises. You conduct a poll of 500 randomly selected city residents, asking them if they own an automobile. 280 say they do own an automobile, and 220 say they do not.
16. Find the sample proportion and sample standard deviation for this data.
17. What is the 95% two-sided confidence interval? Round to four decimal places.
18. Calculate the 90% confidence interval. Round to four decimal places.
19. Calculate the 99% confidence interval. Round to four decimal places.
Use the following information to answer the next three exercises. You are planning to conduct a poll of community members age 65 and older, to determine how many own mobile phones. You want to produce an estimate whose 95% confidence interval will be within four percentage points (plus or minus) the true population proportion. Use an estimated population proportion of 0.5.
20. What sample size do you need?
21. Suppose you knew from prior research that the population proportion was 0.6. What sample size would you need?
22. Suppose you wanted a 95% confidence interval within three percentage points of the population. Assume the population proportion is 0.5. What sample size do you need?
23. In your state, 58 percent of registered voters in a community are registered as Republicans. You want to conduct a study to see if this also holds up in your community. State the null and alternative hypotheses to test this.
24. You believe that at least 58 percent of registered voters in a community are registered as Republicans. State the null and alternative hypotheses to test this.
25. The mean household value in a city is $268,000. You believe that the mean household value in a particular neighborhood is lower than the city average. Write the null and alternative hypotheses to test this.
26. State the appropriate alternative hypothesis to this null hypothesis: H0: μ = 107
27. State the appropriate alternative hypothesis to this null hypothesis: H0: p < 0.25
28. If you reject H0 when H0 is correct, what type of error is this?
29. If you fail to reject H0 when H0 is false, what type of error is this?
30. What is the relationship between the Type II error and the power of a test?
31. A new blood test is being developed to screen patients for cancer. Positive results are followed up by a more accurate (and expensive) test. It is assumed that the patient does not have cancer. Describe the null hypothesis, the Type I and Type II errors for this situation, and explain which type of error is more serious.
32. Explain in words what it means that a screening test for TB has an α level of 0.10. The null hypothesis is that the patient does not have TB.
33. Explain in words what it means that a screening test for TB has a β level of 0.20. The null hypothesis is that the patient does not have TB.
34. Explain in words what it means that a screening test for TB has a power of 0.80.
35. If you are conducting a hypothesis test of a single population mean, and you do not know the population variance, what test will you use if the sample size is 10 and the population is normal?
36. If you are conducting a hypothesis test of a single population mean, and you know the population variance, what test will you use?
37. If you are conducting a hypothesis test of a single population proportion, with np and nq greater than or equal to five, what test will you use, and with what parameters?
38. Published information indicates that, on average, college students spend less than 20 hours studying per week. You draw a sample of 25 students from your college, and find the sample mean to be 18.5 hours, with a standard deviation of 1.5 hours. What distribution will you use to test whether study habits at your college are the same as the national average, and why?
39. A published study says that 95 percent of American children are vaccinated against measles, with a standard deviation of 1.5 percent. You draw a sample of 100 children from your community and check their vaccination records, to see if the vaccination rate in your community is the same as the national average. What distribution will you use for this test, and why?
40. You are conducting a study with an α level of 0.05. If you get a result with a p-value of 0.07, what will be your decision?
41. You are conducting a study with α = 0.01. If you get a result with a p-value of 0.006, what will be your decision?
Use the following information to answer the next five exercises. According to the World Health Organization, the average height of a one-year-old child is 29”. You believe children with a particular disease are smaller than average, so you draw a sample of 20 children with this disease and find a mean height of 27.5” and a sample standard deviation of 1.5”.
42. What are the null and alternative hypotheses for this study?
43. What distribution will you use to test your hypothesis, and why?
44. What is the test statistic and the p-value?
45. Based on your sample results, what is your decision?
46. Suppose the mean for your sample was 25.0. Redo the calculations and describe what your decision would be.
47. You conduct a study using α = 0.05. What is the level of significance for this study?
48. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses:
49. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses:
Use the following information to answer the next three exercises. Nationally, 80 percent of adults own an automobile. You are interested in whether the same proportion in your community own cars. You draw a sample of 100 and find that 75 percent own cars.
50. What are the null and alternative hypotheses for this study?
51. What test will you use, and why?
52. You conduct a poll of political opinions, interviewing both members of 50 married couples. Are the groups in this study independent or matched?
53. You are testing a new drug to treat insomnia. You randomly assign 80 volunteer subjects to either the experimental (new drug) or control (standard treatment) conditions. Are the groups in this study independent or matched?
54. You are investigating the effectiveness of a new math textbook for high school students. You administer a pretest to a group of students at the beginning of the semester, and a posttest at the end of a year’s instruction using this textbook, and compare the results. Are the groups in this study independent or matched?
Use the following information to answer the next two exercises. You are conducting a study of the difference in time at two colleges for undergraduate degree completion. At College A, students take an average of 4.8 years to complete an undergraduate degree, while at College B, they take an average of 4.2 years. The pooled standard deviation for this data is 1.6 years
55. Calculate Cohen’s d and interpret it.
56. Suppose the mean time to earn an undergraduate degree at College A was 5.2 years. Calculate the effect size and interpret it.
57. You conduct an independent-samples t-test with sample size ten in each of two groups. If you are conducting a two-tailed hypothesis test with α = 0.01, what p-values will cause you to reject the null hypothesis?
58. You conduct an independent samples t-test with sample size 15 in each group, with the following hypotheses:
Use the following information to answer the next six exercises. College students in the sciences often complain that they must spend more on textbooks each semester than students in the humanities. To test this, you draw random samples of 50 science and 50 humanities students from your college, and record how much each spent last semester on textbooks. Consider the science students to be group one, and the humanities students to be group two.
59. What is the random variable for this study?
60. What are the null and alternative hypotheses for this study?
61. If the 50 science students spent an average of $530 with a sample standard deviation of $20 and the 50 humanities students spent an average of $380 with a sample standard deviation of $15, would you not reject or reject the null hypothesis? Use an alpha level of 0.05. What is your conclusion?
62. What would be your decision, if you were using α = 0.01?
Use the information to answer the next six exercises. You want to know if proportion of homes with cable television service differs between Community A and Community B. To test this, you draw a random sample of 100 for each and record whether they have cable service.
63. What are the null and alternative hypotheses for this study
64. If 65 households in Community A have cable service, and 78 households in community B, what is the pooled proportion?
65. At α = 0.03, will you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed.
66. Using an alpha value of 0.01, would you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed.
Use the following information to answer the next five exercises. You are interested in whether a particular exercise program helps people lose weight. You conduct a study in which you weigh the participants at the start of the study, and again at the conclusion, after they have participated in the exercise program for six months. You compare the results using a matched-pairs t-test, in which the data is {weight at conclusion – weight at start}. You believe that, on average, the participants will have lost weight after six months on the exercise program.
67. What are the null and alternative hypotheses for this study?
68. Calculate the test statistic, assuming that = –5, sd = 6, and n = 30 (pairs).
69. What are the degrees of freedom for this statistic?
70. Using α = 0.05, what is your decision regarding the effectiveness of this program in causing weight loss? What is the conclusion?
71. What would it mean if the t-statistic had been 4.56, and what would have been your decision in that case?
72. What is the mean and standard deviation for a chi-square distribution with 20 degrees of freedom?
Use the following information to answer the next four exercises. Nationally, about 66 percent of high school graduates enroll in higher education. You perform a chi-square goodness of fit test to see if this same proportion applies to your high school’s most recent graduating class of 200. Your null hypothesis is that the national distribution also applies to your high school.
73. What are the expected numbers of students from your high school graduating class enrolled and not enrolled in higher education?
74. Fill out the rest of this table.
| Observed (O) | Expected (E) | O – E | (O – E)2 | ||
|---|---|---|---|---|---|
| Enrolled | 145 | ||||
| Not enrolled | 55 |
75. What are the degrees of freedom for this chi-square test?
76. What is the chi-square test statistic and the p-value. At the 5% significance level, what do you conclude?
77. For a chi-square distribution with 92 degrees of freedom, the curve _____________.
78. For a chi-square distribution with five degrees of freedom, the curve is ______________.
Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for freshman and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing.
79. Compute the expected values for the cells.
| Cell = Yes | Cell = No | |
|---|---|---|
| Freshman | 100 | 150 |
| Senior | 200 | 50 |
80. Compute for each cell, where O = observed and E = expected.
81. What is the chi-square statistic and degrees of freedom for this study?
82. At the α = 0.5 significance level, what is your decision regarding the null hypothesis?
83. You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test?
84. A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level.
| Ethnic Group \ Response Type | Favor | Oppose | No Opinion | Row Total |
|---|---|---|---|---|
| White / Non-Hispanic | 234 | 433 | 43 | 710 |
| Latino | 147 | 106 | 19 | 272 |
| African American | 24 | 41 | 6 | 71 |
| Asian American | 54 | 48 | 16 | 118 |
| Column Total | 459 | 628 | 84 | 1171 |
85. In a test of homogeneity, what must be true about the expected value of each cell?
86. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence?
87. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity?
88. A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this?
1.
2. normal
3. 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution.
4. z-score = 1.96;
5. 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding.
6. The z-value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085.
7. The standard error of measurement is:
8. The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution.
9. The t-distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”).
10. Both distributions are symmetrical and centered at zero.
11. df = n – 1 = 20 – 1 = 19
12. You can get the t-value from a probability table or a calculator. In this case, for a t-distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e.,
13.
14. The calculator function is invT(0.995, 19).
15. df = n – 1 = 30 – 1 = 29.
16.
17. Because you are using the normal approximation to the binomial, .
18.
19.
20. EBP = 0.04 (because 4% = 0.04)
21.
22.
23. H0: p = 0.58
24. H0: p ≥ 0.58
25. H0: μ ≥ $268,000
26. Ha: μ ≠ 107
27. Ha: p ≥ 0.25
28. a Type I error
29. a Type II error
30. Power = 1 – β = 1 – P(Type II error).
31. The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment.
32. The screening test has a ten percent probability of a Type I error, meaning that ten percent of the time, it will detect TB when it is not present.
33. The screening test has a 20 percent probability of a Type II error, meaning that 20 percent of the time, it will fail to detect TB when it is in fact present.
34. Eighty percent of the time, the screening test will detect TB when it is actually present.
35. The Student’s t-test.
36. The normal distribution or z-test.
37. The normal distribution with μ = p and σ =
38. t24. You use the t-distribution because you don’t know the population standard deviation, and the degrees of freedom are 24 because df = n – 1.
39.
40. Fail to reject the null hypothesis, because α ≤ p
41. Reject the null hypothesis, because α ≥ p.
42. H0: μ ≥ 29.0”
43. t19. Because you do not know the population standard deviation, use the t-distribution. The degrees of freedom are 19, because df = n – 1.
44. The test statistic is –4.4721 and the p-value is 0.00013 using the calculator function TTEST.
45. With α = 0.05, reject the null hypothesis.
46. With α = 0.05, the p-value is almost zero using the calculator function TTEST so reject the null hypothesis.
47. The level of significance is five percent.
48. two-tailed
49. one-tailed
50. H0: p = 0.8
51. You will use the normal test for a single population proportion because np and nq are both greater than five.
52. They are matched (paired), because you interviewed married couples.
53. They are independent, because participants were assigned at random to the groups.
54. They are matched (paired), because you collected data twice from each individual.
55.
56.
57. p-value < 0.01.
58. You will only reject the null hypothesis if you get a value significantly below the hypothesized mean of 110.
59. , i.e., the mean difference in amount spent on textbooks for the two groups.
60. H0: ≤ 0
61. Using the calculator function 2-SampTtest, reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students.
62. Using the calculator function 2-SampTtest, reject the null hypothesis. At the 1% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students.
63. H0: pA = pB
64.
65. Using the calculator function 2-PropZTest, the p-value = 0.0417. Reject the null hypothesis. At the 3% significance level, here is sufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.
66. Using the calculator function 2-PropZTest, the p-value = 0.0417. Do not reject the null hypothesis. At the 1% significance level, there is insufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.
67. H0:
68. t = – 4.5644
69. df = 30 – 1 = 29.
70. Using the calculator function TTEST, the p-value = 0.00004 so reject the null hypothesis. At the 5% level, there is sufficient evidence to conclude that the participants lost weight, on average.
71. A positive t-statistic would mean that participants, on average, gained weight over the six months.
72. μ = df = 20
73. Enrolled = 200(0.66) = 132. Not enrolled = 200(0.34) = 68
74.
| Observed (O) | Expected (E) | O – E | (O – E)2 | ||
|---|---|---|---|---|---|
| Enrolled | 145 | 132 | 145 – 132 = 13 | 169 | |
| Not enrolled | 55 | 68 | 55 – 68 = –13 | 169 |
75. df = n – 1 = 2 – 1 = 1.
76. Using the calculator function Chi-square GOF – Test (in STAT TESTS), the test statistic is 3.7656 and the p-value is 0.0523. Do not reject the null hypothesis. At the 5% significance level, there is insufficient evidence to conclude that high school most recent graduating class distribution of enrolled and not enrolled does not fit that of the national distribution.
77. approximates the normal
78. skewed right
79.
| Cell = Yes | Cell = No | Total | |
|---|---|---|---|
| Freshman | 250 | ||
| Senior | 250 | ||
| Total | 300 | 200 | 500 |
80.
81. Chi-square = 16.67 + 25 + 16.67 + 25 = 83.34.
82. p-value = P(Chi-square, 83.34) = 0
83. The table has five rows and two columns. df = (r – 1)(c – 1) = (4)(1) = 4.
84. Using the calculator function (STAT TESTS) Chi-square Test, the p-value = 0. Reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the poll responses independent of the participants’ ethnic group.
85. The expected value of each cell must be at least five.
86. H0: The variables are independent.
87. H0: The populations have the same distribution.
88. H0: σ2 ≤ 5
1. Which of the following equations is/are linear?
2. To complete a painting job requires four hours setup time plus one hour per 1,000 square feet. How would you express this information in a linear equation?
3. A statistics instructor is paid a per-class fee of $2,000 plus $100 for each student in the class. How would you express this information in a linear equation?
4. A tutoring school requires students to pay a one-time enrollment fee of $500 plus tuition of $3,000 per year. Express this information in an equation.
Use the following information to answer the next four exercises. For the labor costs of doing repairs, an auto mechanic charges a flat fee of $75 per car, plus an hourly rate of $55.
5. What are the independent and dependent variables for this situation?
6. Write the equation and identify the slope and intercept.
7. What is the labor charge for a job that takes 3.5 hours to complete?
8. One job takes 2.4 hours to complete, while another takes 6.3 hours. What is the difference in labor costs for these two jobs?
9. Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression.
10. Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression.
11. Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression.
12. Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression.
Use the following information to answer the next four exercises. Height (in inches) and weight (In pounds) in a sample of college freshman men have a linear relationship with the following summary statistics:
13. What is the value of the slope?
14. What is the value of the y intercept?
15. Write the regression equation predicting weight from height in this data set, and calculate the predicted weight for someone 68 inches tall.
16. The correlation between body weight and fuel efficiency (measured as miles per gallon) for a sample of 2,012 model cars is –0.56. Calculate the coefficient of determination for this data and explain what it means.
17. The correlation between high school GPA and freshman college GPA for a sample of 200 university students is 0.32. How much variation in freshman college GPA is not explained by high school GPA?
18. Rounded to two decimal places what correlation between two variables is necessary to have a coefficient of determination of at least 0.50?
19. Write the null and alternative hypotheses for a study to determine if two variables are significantly correlated.
20. In a sample of 30 cases, two variables have a correlation of 0.33. Do a t-test to see if this result is significant at the α = 0.05 level. Use the formula:
21. In a sample of 25 cases, two variables have a correlation of 0.45. Do a t-test to see if this result is significant at the α = 0.05 level. Use the formula:
Use the following information to answer the next two exercises. A study relating the grams of potassium (Y) to the grams of fiber (X) per serving in enriched flour products (bread, rolls, etc.) produced the equation:
22. For a product with five grams of fiber per serving, what are the expected grams of potassium per serving?
23. Comparing two products, one with three grams of fiber per serving and one with six grams of fiber per serving, what is the expected difference in grams of potassium per serving?
24. In the context of regression analysis, what is the definition of an outlier, and what is a rule of thumb to evaluate if a given value in a data set is an outlier?
25. In the context of regression analysis, what is the definition of an influential point, and how does an influential point differ from an outlier?
26. The least squares regression line for a data set is and the standard deviation of the residuals is 0.4. Does a case with the values x = 2, y = 6.2 qualify as an outlier?
27. The least squares regression line for a data set is and the standard deviation of the residuals is 0.13. Does a case with the values x = 4.1, y = 2.34 qualify as an outlier?
28. What are the five basic assumptions to be met if you want to do a one-way ANOVA?
29. You are conducting a one-way ANOVA comparing the effectiveness of four drugs in lowering blood pressure in hypertensive patients. What are the null and alternative hypotheses for this study?
30. What is the primary difference between the independent samples t-test and one-way ANOVA?
31. You are comparing the results of three methods of teaching geometry to high school students. The final exam scores X1, X2, X3, for the samples taught by the different methods have the following distributions:
32. You conduct a study comparing the effectiveness of four types of fertilizer to increase crop yield on wheat farms. When examining the sample results, you find that two of the samples have an approximately normal distribution, and two have an approximately uniform distribution. Is this a violation of the assumptions for conducting a one-way ANOVA?
Use the following information to answer the next seven exercises. You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain among calves whose feed includes one of the supplements. You have four groups of 30 calves (one is a control group receiving the usual feed, but no supplement). You will conduct a one-way ANOVA after one year to see if there are difference in the mean weight for the four groups.
33. What is SSwithin in this experiment, and what does it mean?
34. What is SSbetween in this experiment, and what does it mean?
35. What are k and i for this experiment?
36. If SSwithin = 374.5 and SStotal = 621.4 for this data, what is SSbetween?
37. What are MSbetween, and MSwithin, for this experiment?
38. What is the F Statistic for this data?
39. If there had been 35 calves in each group, instead of 30, with the sums of squares remaining the same, would the F Statistic be larger or smaller?
40. Which of the following numbers are possible F Statistics?
41. Histograms F1 and F2 below display the distribution of cases from samples from two populations, one distributed F3,15 and one distributed F5,500. Which sample came from which population?
42. The F Statistic from an experiment with k = 3 and n = 50 is 3.67. At α = 0.05, will you reject the null hypothesis?
43. The F Statistic from an experiment with k = 4 and n = 100 is 4.72. At α = 0.01, will you reject the null hypothesis?
44. What assumptions must be met to perform the F test of two variances?
45. You believe there is greater variance in grades given by the math department at your university than in the English department. You collect all the grades for undergraduate classes in the two departments for a semester, and compute the variance of each, and conduct an F test of two variances. What are the null and alternative hypotheses for this study?
1. e. A, B, and C.
2. Let y = the total number of hours required, and x the square footage, measured in units of 1,000. The equation is: y = x + 4
3. Let y = the total payment, and x the number of students in a class. The equation is: y = 100(x) + 2,000
4. Let y = the total cost of attendance, and x the number of years enrolled. The equation is: y = 3,000(x) + 500
5. The independent variable is the hours worked on a car. The dependent variable is the total labor charges to fix a car.
6. Let y = the total charge, and x the number of hours required. The equation is: y = 55x + 75
7. y = 55(3.5) + 75 = 267.50
8. Because the intercept is included in both equations, while you are only interested in the difference in costs, you do not need to include the intercept in the solution. The difference in number of hours required is: 6.3 – 2.4 = 3.9.
9. The X and Y variables have a strong linear relationship. These variables would be good candidates for analysis with linear regression.
10. The X and Y variables have a strong negative linear relationship. These variables would be good candidates for analysis with linear regression.
11. There is no clear linear relationship between the X and Y variables, so they are not good candidates for linear regression.
12. The X and Y variables have a strong positive relationship, but it is curvilinear rather than linear. These variables are not good candidates for linear regression.
13.
14.
15.
16. The coefficient of determination is the square of the correlation, or r2.
17. The coefficient of determination = 0.322 = 0.1024. This is the amount of variation in freshman college GPA that can be explained by high school GPA. The amount that cannot be explained is 1 – 0.1024 = 0.8976 ≈ 0.90. So about 90 percent of variance in freshman college GPA in this data is not explained by high school GPA.
18.
19. H0: ρ = 0
20.
21.
22.
23. Because the intercept appears in both predicted values, you can ignore it in calculating a predicted difference score. The difference in grams of fiber per serving is 6 – 3 = 3 and the predicted difference in grams of potassium per serving is (16)(3) = 48.
24. An outlier is an observed value that is far from the least squares regression line. A rule of thumb is that a point more than two standard deviations of the residuals from its predicted value on the least squares regression line is an outlier.
25. An influential point is an observed value in a data set that is far from other points in the data set, in a horizontal direction. Unlike an outlier, an influential point is determined by its relationship with other values in the data set, not by its relationship to the regression line.
26. The predicted value for y is: . The value of 6.2 is less than two standard deviations from the predicted value, so it does not qualify as an outlier.
27. The predicted value for y is: = 2.3 – 0.1(4.1) = 1.89. The value of 2.32 is more than two standard deviations from the predicted value, so it qualifies as an outlier.
28.
29. H0: μ1 = μ2 = μ3 = μ4
30. The independent samples t-test can only compare means from two groups, while one-way ANOVA can compare means of more than two groups.
31. Each sample appears to have been drawn from a normally distributed populations, the factor is a categorical variable (method), the outcome is a numerical variable (test score), and you were told the samples were independent and randomly selected, so those requirements are met. However, each sample has a different standard deviation, and this suggests that the populations from which they were drawn also have different standard deviations, which is a violation of an assumption for one-way ANOVA. Further statistical testing will be necessary to test the assumption of equal variance before proceeding with the analysis.
32. One of the assumptions for a one-way ANOVA is that the samples are drawn from normally distributed populations. Since two of your samples have an approximately uniform distribution, this casts doubt on whether this assumption has been met. Further statistical testing will be necessary to determine if you can proceed with the analysis.
33. SSwithin is the sum of squares within groups, representing the variation in outcome that cannot be attributed to the different feed supplements, but due to individual or chance factors among the calves in each group.
34. SSbetween is the sum of squares between groups, representing the variation in outcome that can be attributed to the different feed supplements.
35. k = the number of groups = 4
36. SStotal = SSwithin + SSbetween so SSbetween = SStotal – SSwithin
37. The mean squares in an ANOVA are found by dividing each sum of squares by its respective degrees of freedom (df).
38.
39. It would be larger, because you would be dividing by a smaller number. The value of MSbetween would not change with a change of sample size, but the value of MSwithin would be smaller, because you would be dividing by a larger number (dfwithin would be 136, not 116). Dividing a constant by a smaller number produces a larger result.
40. All but choice c, –3.61. F Statistics are always greater than or equal to 0.
41. As the degrees of freedom increase in an F distribution, the distribution becomes more nearly normal. Histogram F2 is closer to a normal distribution than histogram F1, so the sample displayed in histogram F1 was drawn from the F3,15 population, and the sample displayed in histogram F2 was drawn from the F5,500 population.
42. Using the calculator function Fcdf, p-value = Fcdf(3.67, 1E, 3,50) = 0.0182. Reject the null hypothesis.
43. Using the calculator function Fcdf, p-value = Fcdf(4.72, 1E, 4, 100) = 0.0016 Reject the null hypothesis.
44. The samples must be drawn from populations that are normally distributed, and must be drawn from independent populations.
45. Let = variance in math grades, and = variance in English grades.
Use the following information to answer the next two exercises: An experiment consists of tossing two, 12-sided dice (the numbers 1–12 are printed on the sides of each die).
1. Events A and B are:
2. Find P(A|B).
3. Which of the following are TRUE when we perform a hypothesis test on matched or paired samples?
| Light colors | Dark colors | Vibrant colors | |
|---|---|---|---|
| Female | 20 | 22 | 28 |
| Male | 10 | 30 | 8 |
4. Find the probability that a randomly chosen student is male or has a bedroom painted with light colors.
5. Find the probability that a randomly chosen student is male given the student’s bedroom is painted with dark colors.
| x | P (x) |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 |
6. Find the probability that a teenager is reminded two times.
7. Find the expected number of times a teenager is reminded to do his or her chores.
8. For every 22 cars, how many would you expect to be parked crookedly, on average?
9. What is the probability that at least ten of the 22 cars are parked crookedly.
10. Using a sample of 15 Stanford-Binet IQ scores, we wish to conduct a hypothesis test. Our claim is that the mean IQ score on the Stanford-Binet IQ test is more than 100. It is known that the standard deviation of all Stanford-Binet IQ scores is 15 points. The correct distribution to use for the hypothesis test is:
11. If you were to conduct an appropriate hypothesis test, the alternate hypothesis would be:
12. The Type I error is to:
13. The correct decision is to:
Kia, Alejandra, and Iris are runners on the track teams at three different schools. Their running times, in minutes, and the statistics for the track teams at their respective schools, for a one mile run, are given in the table below:
| Running Time | School Average Running Time | School Standard Deviation | |
|---|---|---|---|
| Kia | 4.9 | 5.2 | 0.15 |
| Alejandra | 4.2 | 4.6 | 0.25 |
| Iris | 4.5 | 4.9 | 0.12 |
14. Which student is the BEST when compared to the other runners at her school?
15. The correct distribution to use for the hypothesis test is:
16. The hypothesis test:
17. Sara, a statistics student, wanted to determine the mean number of books that college professors have in their office. She randomly selected two buildings on campus and asked each professor in the selected buildings how many books are in his or her office. Sara surveyed 25 professors. The type of sampling selected is
18. A clothing store would use which measure of the center of data when placing orders for the typical "middle" customer?
19. In a hypothesis test, the p-value is
Use the following information to answer the next three exercises: A community college offers classes 6 days a week: Monday through Saturday. Maria conducted a study of the students in her classes to determine how many days per week the students who are in her classes come to campus for classes. In each of her 5 classes she randomly selected 10 students and asked them how many days they come to campus for classes. Each of her classes are the same size. The results of her survey are summarized in Table 13.45.
| Number of Days on Campus | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 1 | 2 | ||
| 2 | 12 | .24 | |
| 3 | 10 | .20 | |
| 4 | .98 | ||
| 5 | 0 | ||
| 6 | 1 | .02 | 1.00 |
20. Combined with convenience sampling, what other sampling technique did Maria use?
21. How many students come to campus for classes four days a week?
22. What is the 60th percentile for the this data?
Use the following information to answer the next two exercises: The following data are the results of a random survey of 110 Reservists called to active duty to increase security at California airports.
| Number of Dependents | Frequency |
|---|---|
| 0 | 11 |
| 1 | 27 |
| 2 | 33 |
| 3 | 20 |
| 4 | 19 |
23. Construct a 95% confidence interval for the true population mean number of dependents of Reservists called to active duty to increase security at California airports.
24. The 95% confidence interval above means:
25. X ~ U(4, 10). Find the 30th percentile.
26. If X ~ Exp(0.8), then P(x < μ) = __________
27. The lifetime of a computer circuit board is normally distributed with a mean of 2,500 hours and a standard deviation of 60 hours. What is the probability that a randomly chosen board will last at most 2,560 hours?
28. A survey of 123 reservists called to active duty as a result of the September 11, 2001, attacks was conducted to determine the proportion that were married. Eighty-six reported being married. Construct a 98% confidence interval for the true population proportion of reservists called to active duty that are married.
29. Winning times in 26 mile marathons run by world class runners average 145 minutes with a standard deviation of 14 minutes. A sample of the last ten marathon winning times is collected. Let x = mean winning times for ten marathons. The distribution for x is:
30. Suppose that Phi Beta Kappa honors the top one percent of college and university seniors. Assume that grade point means (GPA) at a certain college are normally distributed with a 2.5 mean and a standard deviation of 0.5. What would be the minimum GPA needed to become a member of Phi Beta Kappa at that college?
The number of people living on American farms has declined steadily during the 20th century. Here are data on the farm population (in millions of persons) from 1935 to 1980.
| Year | 1935 | 1940 | 1945 | 1950 | 1955 | 1960 | 1965 | 1970 | 1975 | 1980 |
|---|---|---|---|---|---|---|---|---|---|---|
| Population | 32.1 | 30.5 | 24.4 | 23.0 | 19.1 | 15.6 | 12.4 | 9.7 | 8.9 | 7.2 |
31. The linear regression equation is = 1166.93 – 0.5868x. What was the expected farm population (in millions of persons) for 1980?
32. In linear regression, which is the best possible SSE?
33. In regression analysis, if the correlation coefficient is close to one what can be said about the best fit line?
| Female | Male | |
|---|---|---|
| Accounting | 68 | 56 |
| Administration | 91 | 40 |
| Economics | 5 | 6 |
| Finance | 61 | 59 |
34. The distribution for the test is:
35. The expected number of female who choose finance is:
36. The p-value is 0.0127 and the level of significance is 0.05. The conclusion to the test is:
37. An agency reported that the work force nationwide is composed of 10% professional, 10% clerical, 30% skilled, 15% service, and 35% semiskilled laborers. A random sample of 100 San Jose residents indicated 15 professional, 15 clerical, 40 skilled, 10 service, and 20 semiskilled laborers. At α = 0.10 does the work force in San Jose appear to be consistent with the agency report for the nation? Which kind of test is it?
1. b. independent
2. c.
3. b. Two measurements are drawn from the same pair of individuals or objects.
4. b.
5. d.
6. b.
7. b. 2.78
8. a. 8.25
9. c. 0.2870
10. c. Normal
11. d. Ha: pA ≠ pB
12. b. conclude that the pass rate for Math 1A is different than the pass rate for Math 1B when, in fact, the pass rates are the same.
13. b. not reject H0
14. c. Iris
15. c. Student's t
16. b. is left-tailed.
17. c. cluster sampling
18. b. median
19. a. the probability that an outcome of the data will happen purely by chance when the null hypothesis is true.
20. d. stratified
21. b. 25
22. c. 4
23. a. (1.85, 2.32)
24. c. Both above are correct.
25. c. 5.8
26. c. 0.6321
27. a. 0.8413
28. a. (0.6030, 0.7954)
29. a.
30. d. 3.66
31. b. 5.1
32. a. 13.46
33. b. There is a strong linear pattern. Therefore, it is most likely a good model to be used.
34. b. .
35. d. 70
36. b. There is sufficient evidence to conclude that the choice of major and the gender of the student are not independent of each other.
37. a. goodness-of-fit
1. A study was done to determine the proportion of teenagers that own a car. The population proportion of teenagers that own a car is the:
| value | frequency |
|---|---|
| 0 | 1 |
| 1 | 4 |
| 2 | 7 |
| 3 | 9 |
| 6 | 4 |
2. The box plot for the data is:
3. If six were added to each value of the data in the table, the 15th percentile of the new list of values is:
4. What is the probability of both a drought and water rationing occurring?
5. Which of the following is true?
| gender | apple | pumpkin | pecan |
|---|---|---|---|
| female | 40 | 10 | 30 |
| male | 20 | 30 | 10 |
6. Suppose that one individual is randomly chosen. The probability that the person’s favorite pie is apple or the person is male is _____.
7. Suppose H0 is: Favorite pie and gender are independent. The p-value is ______.
8. Which of the following statements is FALSE?
9. Find the probability that at least six adults watch the news at least once per week.
10. The following histogram is most likely to be a result of sampling from which distribution?
11. The ages of campus day and evening students is known to be normally distributed. A sample of six campus day and evening students reported their ages (in years) as: {18, 35, 27, 45, 20, 20}. What is the error bound for the 90% confidence interval of the true average age?
12. If a normally distributed random variable has µ = 0 and σ = 1, then 97.5% of the population values lie above:
13. What is the probability that one customer spends less than $72 in one trip to the supermarket?
14. How much money altogether would you expect the next five customers to spend in one trip to the supermarket (in dollars)?
15. If you want to find the probability that the mean amount of money 50 customers spend in one trip to the supermarket is less than $60, the distribution to use is:
16. What is the probability that a randomly chosen fourth grader takes more than seven minutes to take out the trash?
17. Which graph best shows the probability that a randomly chosen fourth grader takes more than six minutes to take out the trash given that he or she has already taken more than three minutes?
18. We should expect a fourth grader to take how many minutes to take out the trash?
19. What is the 90th percentile of waiting times (in minutes)?
20. The median waiting time (in minutes) for one student is:
21. Find the probability that the average wait time for ten students is at most 5.5 minutes.
22. A sample of 80 software engineers in Silicon Valley is taken and it is found that 20% of them earn approximately $50,000 per year. A point estimate for the true proportion of engineers in Silicon Valley who earn $50,000 per year is:
23. If P(Z < zα) = 0.1587 where Z ~ N(0, 1), then α is equal to:
24. A professor tested 35 students to determine their entering skills. At the end of the term, after completing the course, the same test was administered to the same 35 students to study their improvement. This would be a test of:
A math exam was given to all the third grade children attending ABC School. Two random samples of scores were taken.
| n | s | ||
|---|---|---|---|
| Boys | 55 | 82 | 5 |
| Girls | 60 | 86 | 7 |
25. Which of the following correctly describes the results of a hypothesis test of the claim, “There is a difference between the mean scores obtained by third grade girls and boys at the 5% level of significance”?
26. In a survey of 80 males, 45 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. The correct conclusion is that:
27. From past experience, a statistics teacher has found that the average score on a midterm is 81 with a standard deviation of 5.2. This term, a class of 49 students had a standard deviation of 5 on the midterm. Do the data indicate that we should reject the teacher’s claim that the standard deviation is 5.2? Use α = 0.05.
28. Three loading machines are being compared. Ten samples were taken for each machine. Machine I took an average of 31 minutes to load packages with a standard deviation of two minutes. Machine II took an average of 28 minutes to load packages with a standard deviation of 1.5 minutes. Machine III took an average of 29 minutes to load packages with a standard deviation of one minute. Find the p-value when testing that the average loading times are the same.
| Number of employees x | 650 | 730 | 810 | 900 | 102 | 107 | 1150 |
| Number of bathrooms y | 40 | 50 | 54 | 61 | 82 | 110 | 121 |
29. Is the correlation between the number of employees and the number of bathrooms significant?
30. The linear regression equation is:
31. If a site has 1,150 employees, approximately how many bathrooms should it have?
32. Suppose that a sample of size ten was collected, with = 4.4 and s = 1.4. H0: σ2 = 1.6 vs. Ha: σ2 ≠ 1.6. Which graph best describes the results of the test?
Sixty-four backpackers were asked the number of days since their latest backpacking trip. The number of days is given in Table 13.45:
| # of days | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 9 | 6 | 12 | 7 | 10 | 5 | 10 |
33. Conduct an appropriate test to determine if the distribution is uniform.
34. Which of the following statements is true when using one-way ANOVA?
1. b. parameter.
2. a.
3. c. seven
4. c. 0.02
5. c. none of the above
6. d.
7. a. ≈ 0
8. b. The values for x are: {1, 2, 3,..., 14}
9. c. 0.9417.
10. d. binomial
11. d. 8.7
12. a. –1.96
13. a. 0.6321
14. d. 360
15. b.
16. a.
17. d.
18. b. 5.5
19. d. 6.92
20. a. 5
21. b. 0.8541
22. b. 0.2
23. a. –1.
24. c. matched pairs, dependent groups.
25. d. Reject H0. There is sufficient evidence to conclude that there is a difference in the mean scores.
26. c. there is sufficient evidence to conclude that the proportion for males is higher than the proportion for females.
27. b. no
28. b. p-value is close to 1.
29. b. No
30. c. = 79.96x – 0.0094
31. d. We should not be estimating here.
32. a.
33. a. The p-value is > 0.10. There is insufficient information to conclude that the distribution is not uniform.
34. c. The test is to determine if the different groups have the same means.
The following tables provide lap times from Terri Vogel's log book. Times are recorded in seconds for 2.5-mile laps completed in a series of races and practice runs.
| Lap 1 | Lap 2 | Lap 3 | Lap 4 | Lap 5 | Lap 6 | Lap 7 | |
|---|---|---|---|---|---|---|---|
| Race 1 | 135 | 130 | 131 | 132 | 130 | 131 | 133 |
| Race 2 | 134 | 131 | 131 | 129 | 128 | 128 | 129 |
| Race 3 | 129 | 128 | 127 | 127 | 130 | 127 | 129 |
| Race 4 | 125 | 125 | 126 | 125 | 124 | 125 | 125 |
| Race 5 | 133 | 132 | 132 | 132 | 131 | 130 | 132 |
| Race 6 | 130 | 130 | 130 | 129 | 129 | 130 | 129 |
| Race 7 | 132 | 131 | 133 | 131 | 134 | 134 | 131 |
| Race 8 | 127 | 128 | 127 | 130 | 128 | 126 | 128 |
| Race 9 | 132 | 130 | 127 | 128 | 126 | 127 | 124 |
| Race 10 | 135 | 131 | 131 | 132 | 130 | 131 | 130 |
| Race 11 | 132 | 131 | 132 | 131 | 130 | 129 | 129 |
| Race 12 | 134 | 130 | 130 | 130 | 131 | 130 | 130 |
| Race 13 | 128 | 127 | 128 | 128 | 128 | 129 | 128 |
| Race 14 | 132 | 131 | 131 | 131 | 132 | 130 | 130 |
| Race 15 | 136 | 129 | 129 | 129 | 129 | 129 | 129 |
| Race 16 | 129 | 129 | 129 | 128 | 128 | 129 | 129 |
| Race 17 | 134 | 131 | 132 | 131 | 132 | 132 | 132 |
| Race 18 | 129 | 129 | 130 | 130 | 133 | 133 | 127 |
| Race 19 | 130 | 129 | 129 | 129 | 129 | 129 | 128 |
| Race 20 | 131 | 128 | 130 | 128 | 129 | 130 | 130 |
| Lap 1 | Lap 2 | Lap 3 | Lap 4 | Lap 5 | Lap 6 | Lap 7 | |
|---|---|---|---|---|---|---|---|
| Practice 1 | 142 | 143 | 180 | 137 | 134 | 134 | 172 |
| Practice 2 | 140 | 135 | 134 | 133 | 128 | 128 | 131 |
| Practice 3 | 130 | 133 | 130 | 128 | 135 | 133 | 133 |
| Practice 4 | 141 | 136 | 137 | 136 | 136 | 136 | 145 |
| Practice 5 | 140 | 138 | 136 | 137 | 135 | 134 | 134 |
| Practice 6 | 142 | 142 | 139 | 138 | 129 | 129 | 127 |
| Practice 7 | 139 | 137 | 135 | 135 | 137 | 134 | 135 |
| Practice 8 | 143 | 136 | 134 | 133 | 134 | 133 | 132 |
| Practice 9 | 135 | 134 | 133 | 133 | 132 | 132 | 133 |
| Practice 10 | 131 | 130 | 128 | 129 | 127 | 128 | 127 |
| Practice 11 | 143 | 139 | 139 | 138 | 138 | 137 | 138 |
| Practice 12 | 132 | 133 | 131 | 129 | 128 | 127 | 126 |
| Practice 13 | 149 | 144 | 144 | 139 | 138 | 138 | 137 |
| Practice 14 | 133 | 132 | 137 | 133 | 134 | 130 | 131 |
| Practice 15 | 138 | 136 | 133 | 133 | 132 | 131 | 131 |
The following table lists initial public offering (IPO) stock prices for all 1999 stocks that at least doubled in value during the first day of trading.
| $17.00 | $23.00 | $14.00 | $16.00 | $12.00 | $26.00 |
| $20.00 | $22.00 | $14.00 | $15.00 | $22.00 | $18.00 |
| $18.00 | $21.00 | $21.00 | $19.00 | $15.00 | $21.00 |
| $18.00 | $17.00 | $15.00 | $25.00 | $14.00 | $30.00 |
| $16.00 | $10.00 | $20.00 | $12.00 | $16.00 | $17.44 |
| $16.00 | $14.00 | $15.00 | $20.00 | $20.00 | $16.00 |
| $17.00 | $16.00 | $15.00 | $15.00 | $19.00 | $48.00 |
| $16.00 | $18.00 | $9.00 | $18.00 | $18.00 | $20.00 |
| $8.00 | $20.00 | $17.00 | $14.00 | $11.00 | $16.00 |
| $19.00 | $15.00 | $21.00 | $12.00 | $8.00 | $16.00 |
| $13.00 | $14.00 | $15.00 | $14.00 | $13.41 | $28.00 |
| $21.00 | $17.00 | $28.00 | $17.00 | $19.00 | $16.00 |
| $17.00 | $19.00 | $18.00 | $17.00 | $15.00 | |
| $14.00 | $21.00 | $12.00 | $18.00 | $24.00 | |
| $15.00 | $23.00 | $14.00 | $16.00 | $12.00 | |
| $24.00 | $20.00 | $14.00 | $14.00 | $15.00 | |
| $14.00 | $19.00 | $16.00 | $38.00 | $20.00 | |
| $24.00 | $16.00 | $8.00 | $18.00 | $17.00 | |
| $16.00 | $15.00 | $7.00 | $19.00 | $12.00 | |
| $8.00 | $23.00 | $12.00 | $18.00 | $20.00 | |
| $21.00 | $34.00 | $16.00 | $26.00 | $14.00 |
Data compiled by Jay R. Ritter of University of Florida using data from Securities Data Co. and Bloomberg.
As you complete each task below, check it off. Answer all questions in your summary.
Here are two examples, but you may NOT use them: number of M&M's per bag, number of pencils students have in their backpacks.
____ What value is 1.5 standard deviations below the mean?
You need to turn in the following typed and stapled packet, with pages in the following order:
As you complete each task below, check it off. Answer all questions in your summary.
____ Decide what continuous data you are going to study. (Here are two examples, but you may NOT use them: the amount of money a student spent on college supplies this term, or the length of time distance telephone call lasts.)
____ Suppose that X followed the following theoretical distributions. Set up each distribution using the appropriate information from your data.
You should have one page for the uniform distribution, one page for the exponential distribution, and one page for the normal distribution.
____ State the distribution: X ~ _________
______ From your original data (before ordering), use a random number generator to pick 40 samples of size five. For each sample, calculate the average.
You need to turn in the following typed and stapled packet, with pages in the following order:
As you complete each task, check it off. Answer all questions in your summary.
Turn in the following typed (12 point) and stapled packet for your final project:
Introduction____State the bivariate data your group is going to study.
Here are two examples, but you may NOT use them: height vs. weight and age vs. running distance.
Analysis ____On a separate sheet of paper construct a scatter plot of the data. Label and scale both axes.
In this section, you will use the data for ONE variable only. Pick the variable that is more interesting to analyze. For example: if your independent variable is sequential data such as year with 30 years and one piece of data per year, your x-values might be 1971, 1972, 1973, 1974, …, 2000. This would not be interesting to analyze. In that case, choose to use the dependent variable to analyze for this part of the project.
Entire Project, typed and stapled: __________
____ Cover sheet: names, class time, and name of your study
____ Part I: label the sections “Intro” and “Analysis.”
____ Part II:
____ Summary page containing several paragraphs written in complete sentences describing the experiment, including what you studied and how you collected your data. The summary page should also include answers to ALL the questions asked above.
____ All graphs requested in the project
____ All calculations requested to support questions in data
____ Description: what you learned by doing this project, what challenges you had, how you overcame the challenges
Include answers to ALL questions asked, even if not explicitly repeated in the items above.
Class Time: __________________________
Class Time: __________________________
Class Time: __________________________
Class Time: __________________________
| When the English says: | Interpret this as: |
|---|---|
| X is at least 4. | X ≥ 4 |
| The minimum of X is 4. | X ≥ 4 |
| X is no less than 4. | X ≥ 4 |
| X is greater than or equal to 4. | X ≥ 4 |
| X is at most 4. | X ≤ 4 |
| The maximum of X is 4. | X ≤ 4 |
| X is no more than 4. | X ≤ 4 |
| X is less than or equal to 4. | X ≤ 4 |
| X does not exceed 4. | X ≤ 4 |
| X is greater than 4. | X > 4 |
| X is more than 4. | X > 4 |
| X exceeds 4. | X > 4 |
| X is less than 4. | X < 4 |
| There are fewer X than 4. | X < 4 |
| X is 4. | X = 4 |
| X is equal to 4. | X = 4 |
| X is the same as 4. | X = 4 |
| X is not 4. | X ≠ 4 |
| X is not equal to 4. | X ≠ 4 |
| X is not the same as 4. | X ≠ 4 |
| X is different than 4. | X ≠ 4 |
Formula 1: Factorial
Formula 2: Combinations
Formula 3: Binomial Distribution
, for
Formula 4: Geometric Distribution
, for
Formula 5: Hypergeometric Distribution
Formula 6: Poisson Distribution
Formula 7: Uniform Distribution
,
Formula 8: Exponential Distribution
Formula 9: Normal Distribution
,
Formula 10: Gamma Function
for , a nonnegative integer
otherwise:
Formula 11: Student's t-distribution
, = degrees of freedom
Formula 12: Chi-Square Distribution
, , = positive integer and degrees of freedom
Formula 13: F Distribution
degrees of freedom for the numerator
degrees of freedom for the denominator
, , are chi-square
| Chapter (1st used) | Symbol | Spoken | Meaning |
|---|---|---|---|
| Sampling and Data | The square root of | same | |
| Sampling and Data | Pi | 3.14159… (a specific number) | |
| Descriptive Statistics | Q1 | Quartile one | the first quartile |
| Descriptive Statistics | Q2 | Quartile two | the second quartile |
| Descriptive Statistics | Q3 | Quartile three | the third quartile |
| Descriptive Statistics | IQR | interquartile range | Q3 – Q1 = IQR |
| Descriptive Statistics | x-bar | sample mean | |
| Descriptive Statistics | mu | population mean | |
| Descriptive Statistics | s sx sx | s | sample standard deviation |
| Descriptive Statistics | s squared | sample variance | |
| Descriptive Statistics | σx | sigma | population standard deviation |
| Descriptive Statistics | sigma squared | population variance | |
| Descriptive Statistics | capital sigma | sum | |
| Probability Topics | brackets | set notation | |
| Probability Topics | S | sample space | |
| Probability Topics | Event A | event A | |
| Probability Topics | probability of A | probability of A occurring | |
| Probability Topics | probability of A given B | prob. of A occurring given B has occurred | |
| Probability Topics | prob. of A or B | prob. of A or B or both occurring | |
| Probability Topics | prob. of A and B | prob. of both A and B occurring (same time) | |
| Probability Topics | A′ | A-prime, complement of A | complement of A, not A |
| Probability Topics | P(A') | prob. of complement of A | same |
| Probability Topics | G1 | green on first pick | same |
| Probability Topics | P(G1) | prob. of green on first pick | same |
| Discrete Random Variables | prob. distribution function | same | |
| Discrete Random Variables | X | X | the random variable X |
| Discrete Random Variables | X ~ | the distribution of X | same |
| Discrete Random Variables | B | binomial distribution | same |
| Discrete Random Variables | G | geometric distribution | same |
| Discrete Random Variables | H | hypergeometric dist. | same |
| Discrete Random Variables | P | Poisson dist. | same |
| Discrete Random Variables | Lambda | average of Poisson distribution | |
| Discrete Random Variables | greater than or equal to | same | |
| Discrete Random Variables | less than or equal to | same | |
| Discrete Random Variables | = | equal to | same |
| Discrete Random Variables | ≠ | not equal to | same |
| Continuous Random Variables | f(x) | f of x | function of x |
| Continuous Random Variables | prob. density function | same | |
| Continuous Random Variables | U | uniform distribution | same |
| Continuous Random Variables | Exp | exponential distribution | same |
| Continuous Random Variables | k | k | critical value |
| Continuous Random Variables | f(x) = | f of x equals | same |
| Continuous Random Variables | m | m | decay rate (for exp. dist.) |
| The Normal Distribution | N | normal distribution | same |
| The Normal Distribution | z | z-score | same |
| The Normal Distribution | Z | standard normal dist. | same |
| The Central Limit Theorem | CLT | Central Limit Theorem | same |
| The Central Limit Theorem | X-bar | the random variable X-bar | |
| The Central Limit Theorem | mean of X | the average of X | |
| The Central Limit Theorem | mean of X-bar | the average of X-bar | |
| The Central Limit Theorem | standard deviation of X | same | |
| The Central Limit Theorem | standard deviation of X-bar | same | |
| The Central Limit Theorem | sum of X | same | |
| The Central Limit Theorem | sum of x | same | |
| Confidence Intervals | CL | confidence level | same |
| Confidence Intervals | CI | confidence interval | same |
| Confidence Intervals | EBM | error bound for a mean | same |
| Confidence Intervals | EBP | error bound for a proportion | same |
| Confidence Intervals | t | Student's t-distribution | same |
| Confidence Intervals | df | degrees of freedom | same |
| Confidence Intervals | student t with a/2 area in right tail | same | |
| Confidence Intervals | ; | p-prime; p-hat | sample proportion of success |
| Confidence Intervals | ; | q-prime; q-hat | sample proportion of failure |
| Hypothesis Testing | H-naught, H-sub 0 | null hypothesis | |
| Hypothesis Testing | H-a, H-sub a | alternate hypothesis | |
| Hypothesis Testing | H-1, H-sub 1 | alternate hypothesis | |
| Hypothesis Testing | alpha | probability of Type I error | |
| Hypothesis Testing | beta | probability of Type II error | |
| Hypothesis Testing | X1-bar minus X2-bar | difference in sample means | |
| Hypothesis Testing | mu-1 minus mu-2 | difference in population means | |
| Hypothesis Testing | P1-prime minus P2-prime | difference in sample proportions | |
| Hypothesis Testing | p1 minus p2 | difference in population proportions | |
| Chi-Square Distribution | Ky-square | Chi-square | |
| Chi-Square Distribution | Observed | Observed frequency | |
| Chi-Square Distribution | Expected | Expected frequency | |
| Linear Regression and Correlation | y = a + bx | y equals a plus b-x | equation of a line |
| Linear Regression and Correlation | y-hat | estimated value of y | |
| Linear Regression and Correlation | correlation coefficient | same | |
| Linear Regression and Correlation | error | same | |
| Linear Regression and Correlation | SSE | Sum of Squared Errors | same |
| Linear Regression and Correlation | 1.9s | 1.9 times s | cut-off value for outliers |
| F-Distribution and ANOVA | F | F-ratio | F-ratio |
Legend
[ ] represents yellow command or green letter behind a key< > represents items on the screenTo adjust the contrastPress
, then hold
to increase the contrast or
to decrease the contrast.
To capitalize letters and wordsPress
to get one capital letter, or press
, then
to set all button presses to capital letters.
You can return to the top-level button values by pressing
again.
To correct a mistakeIf you hit a wrong button, just hit
and start again.
To write in scientific notationNumbers in scientific notation are expressed on the TI-83, 83+, 84, and 84+ using E notation, such that...
To transfer programs or equations from one calculator to another:Both calculators: Insert your respective end of the link cable cable
and press
, then
[LINK].
Calculator receiving information:
<RECEIVE> Calculator sending information:
<TRANSMIT>.ERROR 35 LINK generally means that the cables have not been inserted far enough.
Both calculators: Insert your respective end of the link cable cable
Both calculators: press
, then
[QUIT] to exit when done.
These directions are for entering data with the built-in statistical program.
| Data | Frequency |
|---|---|
| –2 | 10 |
| –1 | 3 |
| 0 | 4 |
| 1 | 5 |
| 3 | 8 |
To begin:
Turn on the calculator.
Access statistics mode.
Select <4:ClrList> to clear data from lists, if desired.
Enter list [L1] to be cleared.
[L1] ,
Display last instruction.
[ENTRY]
Continue clearing remaining lists in the same fashion, if desired.
[L2] ,
Access statistics mode.
Select <1:Edit . . .>
Enter data. Data values go into [L1]. (You may need to arrow over to [L1]).
Type in a data value and enter it. (For negative numbers, use the negate (-) key at the bottom of the keypad).
In [L2], enter the frequencies for each data value in [L1].
Type in a frequency and enter it. (If a data value appears only once, the frequency is "1").
Access statistics mode.
<CALC>.Access <1:1-var Stats>.
Indicate that the data is in [L1]...
[L1] ,
...and indicate that the frequencies are in [L2].
[L2] ,
We will assume that the data is already entered.
We will construct two histograms with the built-in STATPLOT application. The first way will use the default ZOOM. The second way will involve customizing a new graph.
Access graphing mode.
[STAT PLOT]
Select <1:plot 1> to access plotting - first graph.
Use the arrows navigate go to <ON> to turn on Plot 1.
<ON> ,
<Xlist>.If "L1" is not selected, select it.
[L1] ,
<Freq>.Assign the frequencies to [L2].
[L2] ,
Go back to access other graphs.
[STAT PLOT]
To deselect equations:
Access the list of equations.
Select each equal sign (=).
To clear equations:
Access the list of equations.
Use the arrow keys to navigate to the right of each equal sign (=) and clear them.
To draw default histogram:
Access the ZOOM menu.
Select <9:ZoomStat>.
To draw custom histogram:
To draw box plots:
Access graphing mode.
[STAT PLOT]
Select <1:Plot 1> to access the first graph.
Use the arrows to select <ON> and turn on Plot 1.
Use the arrows to select the box plot picture and enable it.
<Xlist>.If "L1" is not selected, select it.
[L1] ,
<Freq>.Indicate that the frequencies are in [L2].
[L2] ,
Go back to access other graphs.
[STAT PLOT]
View the box plot.
[STAT PLOT]
The following data is real. The percent of declared ethnic minority students at De Anza College for selected years from 1970–1995 was:
| Year | Student Ethnic Minority Percentage |
|---|---|
| 1970 | 14.13 |
| 1973 | 12.27 |
| 1976 | 14.08 |
| 1979 | 18.16 |
| 1982 | 27.64 |
| 1983 | 28.72 |
| 1986 | 31.86 |
| 1989 | 33.14 |
| 1992 | 45.37 |
| 1995 | 53.1 |
The TI-83 has a built-in linear regression feature, which allows the data to be edited.The x-values will be in [L1]; the y-values in [L2].
To enter data and do linear regression:
ON Turns calculator on.
Access graphing mode.
[STAT PLOT]
Turn off all plots.
Access the mode menu.
[STAT PLOT]
Navigate to <Float> and then to the right to <3>.
All numbers will be rounded to three decimal places until changed.
Enter statistics mode and clear lists [L1] and [L2], as describe previously.
Enter editing mode to insert values for x and y.
To display the correlation coefficient:
Access the catalog.
[CATALOG]
Arrow down and select <DiagnosticOn>
Access linear regression.
Select the form of y = a + bx.
LinReg
To see the scatter plot:
Access graphing mode.
[STAT PLOT]
Select <1:plot 1> To access plotting - first graph.
Navigate and select <ON> to turn on Plot 1.
<ON>
Select the scatter plot.
<Xlist>.[L1] is not selected, press
[L1] to select it.
Confirm that the data values are in [L1].
<ON>
<Ylist>.Select that the frequencies are in [L2].
[L2] ,
Go back to access other graphs.
[STAT PLOT]
To see the regression graph:
Access the equation menu. The regression equation will be put into Y1.
Access the vars menu and navigate to <5: Statistics>.
<EQ>.<1: RegEQ> contains the regression equation which will be entered in Y1.
To see the residuals and use them to calculate the critical point for an outlier:
Access the list. RESID will be an item on the menu. Navigate to it.
[LIST], <RESID>
Confirm twice to view the list of residuals. Use the arrows to select them.
Store the residuals in [L3].
[L3] ,
Calculate the . Note that
[L3] ,
Store this value in [L4].
[L4] ,
Calculate the critical value using the equation above.
[V] ,
[LIST]
[L4] ,
[L3] to 7.64. If the absolute value is greater than 7.64, then the (x, y) corresponding point is an outlier. In this case, none of the points is an outlier.To obtain estimates of y for various x-values:There are various ways to determine estimates for "y." One way is to substitute values for "x" in the equation. Another way is to use the
on the graph of the regression line.
Access DISTR (for "Distributions").
For technical assistance, visit the Texas Instruments website at http://www.ti.com and enter your calculator model into the "search" box.
Binomial Distribution
binompdf(n,p,x) corresponds to P(X = x)binomcdf(n,p,x) corresponds to P(X ≤ x)x" parameter.Poisson Distribution
poissonpdf(λ,x) corresponds to P(X = x)poissoncdf(λ,x) corresponds to P(X ≤ x)Continuous Distributions (general)
Normal Distribution
normalpdf(x,μ,σ) yields a probability density function value (only useful to plot the normal curve, in which case "x" is the variable)normalcdf(left bound, right bound, μ, σ) corresponds to P(left bound < X < right bound)normalcdf(left bound, right bound) corresponds to P(left bound < Z < right bound) – standard normalinvNorm(p,μ,σ) yields the critical value, k: P(X < k) = pinvNorm(p) yields the critical value, k: P(Z < k) = p for the standard normalStudent's t-Distribution
tpdf(x,df) yields the probability density function value (only useful to plot the student-t curve, in which case "x" is the variable)tcdf(left bound, right bound, df) corresponds to P(left bound < t < right bound)Chi-square Distribution
Χ2pdf(x,df) yields the probability density function value (only useful to plot the chi2 curve, in which case "x" is the variable)Χ2cdf(left bound, right bound, df) corresponds to P(left bound < Χ2 < right bound)F Distribution
Fpdf(x,dfnum,dfdenom) yields the probability density function value (only useful to plot the F curve, in which case "x" is the variable)Fcdf(left bound,right bound,dfnum,dfdenom) corresponds to P(left bound < F < right bound)Access STAT and TESTS.
For the confidence intervals and hypothesis tests, you may enter the data into the appropriate lists and press DATA to have the calculator find the sample means and standard deviations. Or, you may enter the sample means and sample standard deviations directly by pressing STAT once in the appropriate tests.
Confidence Intervals
ZInterval is the confidence interval for mean when σ is known.TInterval is the confidence interval for mean when σ is unknown; s estimates σ.1-PropZInt is the confidence interval for proportion.The confidence levels should be given as percents (ex. enter "95" or ".95" for a 95% confidence level).
Hypothesis Tests
Z-Test is the hypothesis test for single mean when σ is known.T-Test is the hypothesis test for single mean when σ is unknown; s estimates σ.2-SampZTest is the hypothesis test for two independent means when both σ's are known.2-SampTTest is the hypothesis test for two independent means when both σ's are unknown.1-PropZTest is the hypothesis test for single proportion.2-PropZTest is the hypothesis test for two proportions.Χ2-Test is the hypothesis test for independence.Χ2GOF-Test is the hypothesis test for goodness-of-fit (TI-84+ only).LinRegTTEST is the hypothesis test for Linear Regression (TI-84+ only).Input the null hypothesis value in the row below "Inpt." For a test of a single mean, "μ∅" represents the null hypothesis. For a test of a single proportion, "p∅" represents the null hypothesis. Enter the alternate hypothesis on the bottom row.
The module contains links to government site tables used in statistics.
When you are finished with the table link, use the back button on your browser to return here.